12 Projectile Motion Problem Set
PROJECTILE MOTION | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the range of the projectile is closest to: (m)
A. 87 B. 160 C. 214 D. 471
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PROBLEM 2: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the highest point of the trajectory is closest to:
A. 105 m B. 195 m C. 305 m D. 655 m
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PROBLEM 3: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the velocity of this projectile after 2 seconds in flight is closest to:
A. 8.2 m/s B. 19.6 m/s C. 29.7 m/s D. 53 m/s
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PROBLEM 4: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the angle in which this projectile makes with the horizontal axis after 2 seconds in flight is closest to:
A. 188.4° B. 94.1° C. 7.7° D. 18.4°
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PROBLEM 5: A football is thrown at the line of scrimmage with a velocity of 22 m/s at an angle of 26o above the horizontal toward a receiver standing still 35 yards downfield. If the quarterback launches the ball from a point 8 ft above the ground, the height at which the receiver must catch this ball is closest to: A. 5.9 ft B. 8.7 ft C. 2.65 ft D. 13 ft
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PROBLEM 6: A ball is thrown with a velocity of 7 m/s toward a wall that is 5 meters away. If the ball is propelled from a point that is 3 meters above the ground at an angle of 34o above the horizontal, the height at which the ball will hit the wall is closest to:
A. 2.73 m B. 1.44 m C. 0.27 m D. 0.11 m
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PROBLEM 7: A projectile is launched from a cliff 200 meters above level ground with a launch velocity of 50 m/s and a launch angle of 25o above the horizontal. Assuming negligible air resistance, the range of this projectile before making it impact is closest to: (m)
A. 517 B. 51.7 C. 1,517 D. 767
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PROBLEM 8: A projectile is launched from a cliff 200 meters above level ground with a launch velocity of 50 m/s and a launch angle of 25o above the horizontal. Assuming negligible air resistance, the maximum height of its trajectory is closest to: (m)
A. 223 B. 157 C. 336 D. 761
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PROBLEM 9: A projectile is launched from a cliff 200 meters above level ground with a launch velocity of 50 m/s and a launch angle of 25o above the horizontal. Assuming negligible air resistance, the velocity after 1.25 seconds after launch is closest to: (m/s)
A. 8.1 B. 16.3 C. 46.2 D. 102
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PROJECTILE MOTION | SOLUTIONS
SOLUTION 1: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away.
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We know through our studies in PROJECTILE MOTION, that this projectile will follow some defined TRAJECTORY before impact, such that:
We are being asked to determine the RANGE of this projectile before making an impact on the intended target. Or in other words, HOW FAR will this projectile TRAVEL HORIZONTALLY before smashing its target. This may seem like a complicated endeavor that we are about to embark on, but luckily, the process is clean and easily replicated across a number of scenarios, let’s hash it out. We know right off the bat that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT.
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If we flip over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement, although it may not be readily obvious, its there. We can also set the POSITION of the LAUNCH POINT at any position that is convenient for us, as long as we account for that precise location in our computations moving forward.
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Let’s set our LAUNCH POINT to be located at: (𝑥8 , 𝑦8 ) = (0,0) So we are saying that our cannonball will be launched at the COORDINATES of (0,0). All measurements will be relative to this POINT, as identified by the AXIS we have established:
This gives us the ability to DEFINE the relative DATA right up front, which will be: • 𝑣8 = 30 m/s • 𝑥8 = 0 • 𝑦8 = 0 • 𝜃 = 20° • 𝑔 = 9.81 m/s A
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We can also conclude that based on the information that we are presented, that the VERTICAL DISPLACEMENT, y, at the end of the TRAJECTORY will be: 𝑦 = 43 m − 143 m Or: 𝑦 = −100 m It’s important to NOTE the NEGATIVE SIGN here, the ELEVATION DROPS, therefore, there will be a NEGATIVE DISPLACEMENT experienced VERTICALLY by this PROJECTILE. If this is not picked up, our calculations from the start will be incorrect, affecting everything that comes down the line. With this VERTICAL DISPLACEMENT defined for us, let’s revisit our GENERAL FORMULA given to us in the NCEES Reference Handbook, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Which we have defined: • 𝑣8 = 30 m/s • 𝑦8 = 0 • 𝜃 = 20°
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• 𝑔 = 9.81 m/s A • 𝑦 = −100 m We have everything we need with the exception of the TIME, t, so let’s plug and chug some numbers to quantify this VALUE. Plugging in our data we have: (9.81 m/s A )𝑡 A −100 m = − + (30 m/s) sin 20° 𝑡 + 0 2 Simplifying this expression, and removing the UNITS for the time being, we get: −100 m = −4.91𝑡 A + 10.26𝑡 We now have a QUADRATIC EQUATION that we can solve using the QUADRATIC FORMULA, which in the context of our VARIABLE for TIME, t, is written as: −𝑏 ± 𝑏 A − 4𝑎𝑐 𝑡= 2𝑎 Where: 𝑎 = −4.91 𝑏 = 10.26 𝑐 = 100
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Plugging these VALUES in to QUADRATIC FORMULA, we have: −(10.26) ± (10.26)A − 4(−4.91)(100) 𝑡= 2(−4.91) Which gives us the TWO SOLUTIONS: 𝑡I = −3.59 s 𝑡A = 5.68 s We obviously can’t go “back” in TIME, so the negative solution is irrelevant, telling us that the PROJECTILE is in MOTION for 5.68 SECONDS until it makes an IMPACT at the location of the TARGET. We can now refer back to our GENERAL FORMULA given to us in the NCEES Reference Handbook for the HORIZONTAL DISPLACEMENT, which is: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 Where: • 𝑣8 = 30 m/s • 𝑥8 = 0 • 𝜃 = 20° • 𝑡 = 5.68 s
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We have all the necessary DATA defined and are now at a point of a simple plug and chug of our numbers to quantify the RANGE. Plugging in our data we have: 𝑥 = 30 m/s cos 20° (5.68 s) + 0 Or that: 𝑥 = 160 m The RANGE of the projectile before impact at the location of the target is 160 m. The correct answer choice is B. 160 m
SOLUTION 2: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant.
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In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away.
We know through our studies in PROJECTILE MOTION, that the cannonball will follow a defined TRAJECTORY, and at some defined point, 𝑦JKL , it will be at its PEAK, such that it changes from a direction of traveling upwards to one that is traveling downwards. This PEAK height above the ELEVATION of the TARGET is what we are after:
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To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine when this PEAK occurs and the MEASUREMENT at this POINT, they are: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Our approach will be to take the INITIAL VELOCITY at the LAUNCH POINT, along with its associated GEOMETRY, break it down it to its VERTICAL and HORIZONTAL COMPONENTS, and determine where the VERTICAL VELOCITY is ZERO. By observation, we know that at IMPACT, that this will be the case…but it will also be the case at the TOP of the TRAJECTORY, 𝑦JKL , where the VELOCITY in the VERTICAL DIRECTION transitions from going UPWARD to DOWNWARD. There is an INSTANTANEOUS VELOCITY of ZERO at this particular point, and the TIME it takes to get to that POINT is what we are initially after, so let’s get to work.
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Taking the full TRAJECTORY of the CANNON, let’s hone in on the LAUNCH POINT, such that:
Breaking down the INITIAL VELOCITY, we find that the HORIZONTAL and VERTICAL COMPONENTS will be: 𝑣8L = 𝑣8 cos 𝜃 = (30 m/s) cos 20° 𝑣8M = 𝑣8 sin 𝜃 = (30 m/s) sin 20° Or: 𝑣8L = 28.2 m/s 𝑣8M = 10.26 m/s Now we aren’t concerned with the HORIZONTAL COMPONENT at this point, but we do want to dive in to analyzing how that VERTICAL COMPONENT will be affected.
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As we have established from the NCEES Reference Handbook, the HORIZONTAL COMPONENT of VELOCITY at ANY POINT in the TRAJECTORY can be expressed as: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 Where: 𝑣8 sin 𝜃 = 𝑣8M So: 𝑣M = −𝑔𝑡 + 𝑣8M We know that: 𝑣8M = 10.26 m/s 𝑔 = 9.81 m/s A And that at the PEAK of TRAJECTORY: 𝑣M = 0 m/s Plugging all of these values in to the FORMULA, we get: 0 = −(9.81 m/s A )𝑡 + 10.26 m/s
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Rearranging to isolate and solve for the TIME to this POINT, we get:
𝑡=
10.26 m/s = 1.05 s 9.81 m/s A
Knowing this, we can now refer back to the second formula we pulled from the NCEES Reference Handbook, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Where: • 𝑣8 = 30 m/s • 𝑦8 = 100 m • 𝜃 = 20° • 𝑔 = 9.81 m/s A • 𝑡 = 1.05 s Simply plugging in this DATA, we have: (9.81 m/s A )(1.05 s)A 𝑦=− + (30 m/s) sin 20° (1.05 s) + 100 m 2 Which gives us: 𝑦 = 105 m
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The HIGHEST POINT of the trajectory will be 105 meters above the ELEVATION of the TARGET. The correct answer choice is A. 105 m
SOLUTION 3: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away:
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We know through our studies in PROJECTILE MOTION, that the cannonball will follow a defined TRAJECTORY and at any point in that TRAJECTORY it will have a VELOCITY that can be broken up in to its individual HORIZONTAL and VERTICAL COMPONENTS. Having these COMPONENTS defined will allow for us to find the overall MAGNITUDE of the VELOCITY, which in this problem, is asked to be defined 2 seconds after launch:
To start, let’s flip over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine VELOCITY of the PROJECTILE at any point in its TRAJECTORY, they are: 𝑣L = 𝑣8 cos 𝜃 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃
Where: • 𝑣8 = INITIAL VELOCITY
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• 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME In our problem statement we are given all of this DATA, which is: • 𝑣8 = 30 m/s • 𝜃 = 20° • 𝑔 = 9.81 m/s A • 𝑡 = 2 s Let’s take this information and determine each of our VELOCITY COMPONENTS, starting with our HORIZONTAL COMPONENT, which will be: 𝑣L = (30 m/s) cos 20° Or: 𝑣L = 28.2 m/s Doing the same for our VERTICAL COMPONENT, we have: 𝑣M = −(9.81 m/s A )(2 s) + (30 m/s) sin 20°
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Or: 𝑣M = −9.36 m/s Its important to NOTE here the NEGATIVE SIGN in front of our VERTICAL VELOCITY at this POINT. What this is telling us is that the VERTICAL VELOCITY at this POINT in the TRAJECTORY is going in the OPPOSITE DIRECTION then what we had established as the POSITIVE DIRECTION at the beginning of our problem. In other words, we established that the INITIAL VELOCITY going UPWARDS was POSITIVE…the NEGATIVE sign is telling us that now the VELOCITY of the projectile has it headed DOWNWARDS…for this COMPONENT. You may also notice that there was no TIME VARIABLE tied to the HORIZONTAL COMPONENT of VELOCITY. Let’s recall that when working PROJECTILE MOTION problems, we are analyzing under IDEAL CONDITIONS and the assumption that AIR RESISTANCE is NEGLIGIBLE. The only FORCE of significance that acts on any object, or PARTICLE for that matter, is GRAVITY.
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Gravity will always act VERTICALLY DOWNWARD from the center of the OBJECTS MASS causing a DOWNWARD ACCELERATION back towards the surface of the EARTH, or the realm in which the PARTICLE resides. Due to the objects INERTIA created by the initial projection, and the fact that we are working under the assumption that AIR RESISTANCE is NEGLIGABLE, there is created an instance where there is no need for an external horizontal force to maintain the HORIZONTAL COMPONENT of the objects VELOCITY…it remains CONSTANT. However, the VERTICAL COMPONENT of the objects VELOCITY is under CONSTANT GRAVITATIONAL ACCELERATION meaning that it will reduce over TIME. At this POINT, we now have defined the HORIZONTAL and VERTICAL COMPONENTS of VELOCITY 2 seconds after LAUNCH, which are: 𝑣L = 28.2 m/s 𝑣M = −9.36 m/s Solving for the MAGNITUDE of the VELOCITY, we can use the FORMULA:
𝑣=
𝑣L A + 𝑣M A
Which gives us:
𝑣=
(28.2 m/s)A + (−9.36 m/s)A
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Or: 𝑣 = 29.7 m/s The velocity of this projectile after 2 seconds in flight is 29.7 m/s. The correct answer choice is C. 29.7 m/s
SOLUTION 4: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away.
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We know through our studies in PROJECTILE MOTION, that the cannonball will follow a defined TRAJECTORY and at any point in that TRAJECTORY it will have a VELOCITY that can be broken up in to its individual HORIZONTAL and VERTICAL COMPONENTS. Having these COMPONENTS defined will allow for us to find the ANGLE in which the VELOCITY VECTOR is making with the HORIZONTAL AXIS 2 seconds after launch:
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To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine VELOCITY of the PROJECTILE at any point in its TRAJECTORY, they are: 𝑣L = 𝑣8 cos 𝜃 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃
Where: • 𝑣8 = INITIAL VELOCITY • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME In our problem statement we are given all of this DATA, which is: • 𝑣8 = 30 m/s • 𝜃 = 20° • 𝑔 = 9.81 m/s A • 𝑡 = 2 s Let’s take this information and determine each of our VELOCITY COMPONENTS, starting with our HORIZONTAL COMPONENT, which will be: 𝑣L = (30 m/s) cos 20°
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Or: 𝑣L = 28.2 m/s Doing the same for our VERTICAL COMPONENT, we have: 𝑣M = −(9.81 m/s A )(2 s) + (30 m/s) sin 20° Or: 𝑣M = −9.36 m/s Its important to NOTE here the NEGATIVE SIGN in front of our VERTICAL VELOCITY at this POINT. What this is telling us is that the VERTICAL VELOCITY at this POINT in the TRAJECTORY is going in the OPPOSITE DIRECTION then what we had established as the POSITIVE DIRECTION at the beginning of our problem. In other words, we established that the INITIAL VELOCITY going UPWARDS was POSITIVE…the NEGATIVE sign is telling us that now the VELOCITY is headed DOWNWARDS…for this COMPONENT. You may also notice that there was no TIME VARIABLE tied to the HORIZONTAL COMPONENT of VELOCITY.
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Let’s recall that when working PROJECTILE MOTION problems, we are analyzing under IDEAL CONDITIONS and the assumption that AIR RESISTANCE is NEGLIGIBLE. The only FORCE of significance that acts on any object, or PARTICLE for that matter, is GRAVITY. Gravity will always act VERTICALLY DOWNWARD from the center of the OBJECTS MASS causing a DOWNWARD ACCELERATION back towards the surface of the EARTH, or the realm in which the PARTICLE resides. Due to the objects INERTIA created by the initial projection, and the fact that we are working under the assumption that AIR RESISTANCE is NEGLIGABLE, there is created an instance where there is no need for an external horizontal force to maintain the HORIZONTAL COMPONENT of the objects VELOCITY…it remains CONSTANT. However, the VERTICAL COMPONENT of the objects VELOCITY is under CONSTANT GRAVITATIONAL ACCELERATION meaning that it will reduce over TIME. At this POINT, we now have defined the HORIZONTAL and VERTICAL COMPONENTS of VELOCITY 2 seconds after LAUNCH, which are: 𝑣L = 28.2 m/s 𝑣M = −9.36 m/s
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Referring back to our TRIGONOMETRY RELATIONSHIPS, we know that:
tan(𝜃) =
𝑦 𝑥
Or in terms of the famous adage SOH CAH TOA:
tan(𝜃) =
opposite adjacent
Whichever way you remember best is the one we want you to roll with. Taking our COMPONENTS, we can illustrate the GEOMETRY of the VELOCITY VECTOR at this POINT as:
At this POINT in the TRAJECTORY the projectile is going to the RIGHT and DOWNWARD, meaning that the ANGLE with the HORIZONTAL AXIS will be BELOW the AXIS, as illustrated.
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Our COMPONENTS can then be defined as: Adjacent = 𝑣L = 28.2 m/s Opposite = 𝑣M = 9.36 m/s Plugging these into our TRIGONOMETRIC RELATIONSHIP for the TANGENT, we have:
tan(𝜃) =
9.36 m/s 28.2 m/s
Rearranging to ISOLATE and SOLVE for the ANGLE, we get:
𝜃 = tanUI
9.36 m/s 28.2 m/s
Or: 𝜃 = 18.4° Now the important characteristic to NOTE here is the LOCATION of this ANGLE. Come exam day, it may be measured starting at the POSITIVE X-AXIS and COUNTERCLOCKWISE, making it 341.6°. It may also be presented as an answer option with a NEGATIVE SIGN, indicating its LOCATION relative to the POSITIVE X-AXIS.
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Nonetheless, the ANGLE that the VELOCITY VECTOR of this projectile makes with the HORIZONTAL AXIS after 2 seconds in flight is 18.4°. The correct answer choice is D. 18.40
SOLUTION 5: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario where a quarterback is throwing a football 35 yards downfield to a receiver standing still. Given the information presented to us in the problem statement, we want to know at what elevation the receiver must go to catch this ball at the point he is located. This may seem like a complicated task, but luckily, the process is clean and easily replicated across a number of scenarios, let’s hash it out.
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We know right off the bat that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT. Taking a look at the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement, although it may not be directly apparent, its there.
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We can also set the POSITION of the AXIS in which the LAUNCH POINT is measured to any point that is convenient for us, as long as we account for that precise location in our computations moving forward. Let’s set our LAUNCH AXIS at the feet of the quarterback, and directly below the HORIZONTAL position of the LAUNCH POINT. This makes our COORDINATES then: (𝑥8 , 𝑦8 ) = ( 0, 8 ft) All measurements will be relative to this AXIS. We can now DEFINE the relative DATA presented to us, which is: • 𝑣8 = 22 m/s • 𝑥8 = 0 • 𝑦8 = 8 ft • 𝜃 = 26° • 𝑔 = 9.81 m/s A We also know that: • 𝑥 = 35 yards Various UNITS of measure are being used here, so let’s convert everything in to SI to make sure all calculations moving forward are accurate.
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This will be a point that many students will muddy up a fairly straight forward problem, so keep calm under the pressure and go through your mental checklists to ensure everything is laying out correctly. Converting our HORIZONTAL DISPLACEMENT as well as the VERTICAL LAUNCH POINT, we get: • 𝑥 = 38.3 m • 𝑦8 = 2.43 m The most common CONVERSION FACTORS can be found on PAGE 2 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Everything is now looking good, let’s revisit the formulas we pulled from the PARTICLE KINEMATICS section once again, they are: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
We have most everything we need to just revert to a plug and play problem.
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With the DISPLACEMENT formulas presented to us, let’s create a pair of expressions representing our scenario, this gives us: 38.3 m = (22 m/s) cos 26° 𝑡 + 0 (9.81 m/s A )𝑡 A 𝑦=− + (22 m/s) sin 26° 𝑡 + 2.43 m 2
We are now at a point where we have TWO EQUATIONS and TWO UNKNOWNS…everything is set up perfectly for us to quantify these two values, of which, we are ultimately concerned with defining the HEIGHT in which the receiver must go to catch this ball. With this pair of EQUATIONS set up, we would recommend from here that you learn how to HACK this problem, solving it in a FRACTION of the time using your NCEES APPROVED CALCULATOR. As a student of PREPINEER, you have full access to seeing how this is done on your calculator of choice using the CALCULATOR HACKS that we have developed and archived for your ON-DEMAND TRAINING. But we will still walk through the step by step calculations because we believe in the importance of fully understanding the fundamentals behind all the work that we do, first and foremost, so let’s do just that.
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Simplifying the expressions that we have derived and because we like to work with just numbers, let’s keep in mind the UNITS we are working with, but drop them for now: 38.3 = 19.8𝑡 𝑦 = −4.91𝑡 A + 9.64𝑡
Alright, so those look a lot friendlier to the eye. With only a SINGLE UNKNOWN VARIABLE, let take the FIRST EXPRESSION and rearrange it to ISOLATE and solve for TIME, t. Doing so we get: 𝑡 = 1.94 s Taking this newly defined value for TIME and plugging it in to the SECOND EXPRESSION, we get: 𝑦 = −4.91 1.94 s
A
+ 9.64 1.94 s + 2.43 m
Or that: 𝑦 = 2.65 m So given that a quarterback chucks a football from the line of scrimmage with a velocity of 22 m/s and at an angle of 26o above the horizontal, it must be received at a point that
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is 2.65 meters above the ground…or since the answer options are given in feet, converting back to this UNIT, the reception must be made at a point that is 8.7 feet above the ground. The correct answer choice is B. 8.7 ft
SOLUTION 6: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario where a ball is thrown at a wall that is 5 meters away. Given the information presented to us in the problem statement, we want to know at what elevation the ball will make impact with the wall. By now, it’s clear that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT.
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Turning to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement. We can set the POSITION of the AXIS in which the LAUNCH POINT is measured to any point that is convenient for us, as long as we account for that precise location in our computations moving forward. Let’s set our AXIS at the actual LAUNCH POINT.
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This makes our COORDINATES then: (𝑥8 , 𝑦8 ) = ( 0, 0) All measurements will be relative to this POINT, as identified by the AXIS we have established. This gives us the ability to DEFINE the relative DATA right up front, which will be: • 𝑣8 = 7 m/s • 𝑥8 = 0 • 𝑦8 = 0 • 𝜃 = 34° • 𝑔 = 9.81 m/s A We also know that: • 𝑥 = 5 meters Everything is looking good at this point, so let’s revisit the formulas we pulled from the PARTICLE KINEMATICS section once again, which are: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
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We have most everything we need to just revert to a plug and play problem. With the DISPLACEMENT formulas presented to us, let’s create a pair of expressions representing this scenario, giving us: 5 m = (7 m/s) cos 34° 𝑡 + 0 (9.81 m/s A )𝑡 A 𝑦=− + (7 m/s) sin 34° 𝑡 + 0 2
We are now at a point where we have TWO EQUATIONS and TWO UNKNOWNS…everything is set up perfectly for us to quantify these two values, of which, we are ultimately concerned with defining the HEIGHT in which the ball strikes the wall. With this pair of EQUATIONS set up, we would recommend from here that you learn how to HACK this problem, solving it in a FRACTION of the time using your NCEES APPROVED CALCULATOR. As a student of PREPINEER, you have full access to seeing how this is done on your calculator of choice using the CALCULATOR HACKS that we have developed and archived for your ON-DEMAND TRAINING. But we will still walk through the step by step calculations because we believe in the importance of fully understanding the fundamentals behind all the work that we do, first and foremost, so let’s do just that.
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Simplifying the expressions that we have derived and because we like to work with just numbers, let’s keep in mind the UNITS we are working with, but drop them for now: 5 = 5.8𝑡 𝑦 = −4.91𝑡 A + 3.91𝑡
With only a SINGLE UNKNOWN, take the FIRST EXPRESSION and rearrange it to ISOLATE and solve for the VARIABLE t.
Doing so we get:
𝑡 = .86 s Taking this value for TIME and plugging it in to the SECOND EXPRESSION, we get: 𝑦 = −4.91 . 86
A
+ 3.91(.86)
Or that: 𝑦 = −.27 m Now this answer might surprise you, and certainly would make you doubt the work that you have done up to this point.
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Many will stop, reconsider doing all their calculations over, and go from there…some will just quit and go to the next problem. However, all is well here, all we need to do is realize that the ball is being released at a POINT that is 3 meters above the ground. Our calculations are telling us that the height of the ball will be .27 meters below this starting point, or that: 𝑦 = 3 m − .27 m = 2.73 m The ball will strike at a point that is 2.73 meters above the ground. The correct answer choice is A. 2.73 m
SOLUTION 7: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant.
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In this problem, we are given an illustration and scenario of a projectile being launched from some contraption located 200 meters above the ground, horizontally, towards a target that is some unknown distance away. We are being asked to determine the RANGE of this projectile before making an impact. Or in other words, HOW FAR this projectile will TRAVEL HORIZONTALLY before smashing in to the ground. We inherently know that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT. Flipping back to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT Made with by Prepineer | Prepineer.com
• 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement. We can set the POSITION of the LAUNCH POINT at any point that is convenient for us, as long as we account for that precise location in our computations moving forward. Let’s set our AXIS, relative to our LAUNCH POINT, at the location: (𝑥8 , 𝑦8 ) = ( 0, 200 ) Essentially this would be at the ground level, directly below the LAUNCH POINT…indicating that the PROJECTILE has a starting POSITION that is ZERO UNITS in the HORIZONTAL DIRECTION and 200 UNITS in the VERTICAL DIRECTION. All measurements will be relative to this AXIS. This gives us the ability to DEFINE the relative DATA right up front, which will be: • 𝑣8 = 50 m/s • 𝑥8 = 0 • 𝑦8 = 200 m • 𝜃 = 25° • 𝑔 = 9.81 m/s A
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With this information defined, let’s revisit our GENERAL FORMULA given to us in the NCEES Reference Handbook for the HORIZONTAL DISPLACEMENT, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Of the variables needed, we have: • 𝑣8 = 50 m/s • 𝑦8 = 200 m • 𝜃 = 25° • 𝑔 = 9.81 m/s A We also know that this PROJECTILE will come crashing down at an ELEVATION that is 200 meters below where it started, therefore: 𝑦 = −200 m We have everything we need with the exception of the TIME, t, so let’s plug and chug some numbers to quantify this VALUE. Plugging in our data we have: (9.81 m/s A )𝑡 A −200 m = − + (50 m/s) sin 25° 𝑡 + 200 m 2
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Simplifying this expression, and removing the UNITS for the time being, we get: −400 m = −4.91𝑡 A + 21.1𝑡 We now have a QUADRATIC EQUATION that we can solve using the QUADRATIC FORMULA, which in the context of our VARIABLE for TIME, t, is written as: −𝑏 ± 𝑏 A − 4𝑎𝑐 𝑡= 2𝑎 Where: 𝑎 = −4.91 𝑏 = 21.1 𝑐 = 400 Plugging these VALUES in to QUADRATIC FORMULA, we have: −(21.1) ± (21.1)A − 4(−4.91)(400) 𝑡= 2(−4.91) Which gives us the TWO SOLUTIONS: 𝑡I = −7.13 s 𝑡A = 11.4 s
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We obviously can’t go “back” in TIME, so the negative solution is irrelevant, telling us that the PROJECTILE is in MOTION for 11.4 SECONDS until it makes an IMPACT at the location of the TARGET. We can now refer back to our GENERAL FORMULA given to us in the NCEES Reference Handbook for the HORIZONTAL DISPLACEMENT, which is: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 Which we have defined: • 𝑣8 = 50 m/s • 𝑥8 = 0 • 𝜃 = 25° • 𝑡 = 11.4 s Plugging in our numbers to quantify the RANGE, we have: 𝑥 = (50 m/s) cos 25° (11.4 s) + 0 Or that: 𝑥 = 517 m The RANGE of the projectile before impact at the location of the target is 517 m. The correct answer choice is A. 517 m
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SOLUTION 8: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from some contraption located 200 meters above the ground, horizontally, towards a target that is some unknown distance away. We know through our studies in PROJECTILE MOTION, that the projectile will follow a defined TRAJECTORY, and at some defined point, 𝑦JKL , it will be at its PEAK, such that it changes from a direction of traveling UPWARDS to one that is traveling DOWNWARDS. This PEAK height above the ELEVATION of the TARGET is what we are after.
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To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine when this PEAK occurs and the MEASUREMENT at this POINT, they are: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Our approach will be to take the INITIAL VELOCITY at the LAUNCH POINT, along with its associated GEOMETRY, break it down it to its VERTICAL and HORIZONTAL COMPONENTS, and determine where the VERTICAL VELOCITY is ZERO. By observation, we know that at IMPACT, that this will be the case…but it will also be the case at the TOP of the TRAJECTORY, 𝑦JKL , where the VELOCITY in the VERTICAL DIRECTION transitions from going UPWARD to DOWNWARD. There is an INSTANTANEOUS VELOCITY of ZERO at this particular point, and the TIME it takes to get to that POINT is what we are initially after, so let’s get to work. Breaking down the INITIAL VELOCITY, we find that the HORIZONTAL and VERTICAL COMPONENTS will be: 𝑣8L = 𝑣8 cos 𝜃 = (50 m/s) cos 25° 𝑣8M = 𝑣8 sin 𝜃 = (50 m/s) sin 25°
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Or: 𝑣8L = 45.3 m/s 𝑣8M = 21.1 m/s Now we aren’t concerned with the HORIZONTAL COMPONENT at this point, but we do want to dive in to analyzing how that VERTICAL COMPONENT will be affected. As we have established from the NCEES Reference Handbook, the HORIZONTAL COMPONENT of VELOCITY at ANY POINT in the TRAJECTORY can be expressed as: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 Where: 𝑣8 sin 𝜃 = 𝑣8M So: 𝑣M = −𝑔𝑡 + 𝑣8M We know that: 𝑣8M = 21.1 m/s 𝑔 = 9.81 m/s A
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And that at the PEAK of TRAJECTORY: 𝑣M = 0 m/s Plugging all of these values in to the FORMULA, we get: 0 = −(9.81 m/s A )𝑡 + 21.1 m/s Rearranging to isolate and solve for the TIME to this POINT, we get:
𝑡=
21.1 m/s = 2.15 s 9.81 m/s A
Knowing this, we can now refer back to the second formula we pulled from the NCEES Reference Handbook, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Where we have defined: • 𝑣8 = 50 m/s • 𝑦8 = 200 m • 𝜃 = 25° • 𝑔 = 9.81 m/s A • 𝑡 = 2.15 s
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Simply plugging in this DATA, we have: (9.81 m/s A )(2.15 s)A 𝑦=− + (50 m/s) sin 25° (2.15 s) + 200 m 2 Which gives us: 𝑦 = 223 m The HIGHEST POINT of the trajectory is 223 meters above the ELEVATION of the TARGET. The correct answer choice is A. 223 m
SOLUTION 9: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant.
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In this problem, we are given an illustration and scenario of a projectile being launched from some contraption located 200 meters above the ground, horizontally, towards a target that is some unknown distance away. We know through our studies in PROJECTILE MOTION, that the projectile will follow a defined TRAJECTORY and at any point in that TRAJECTORY it will have a VELOCITY that can be broken up in to its individual HORIZONTAL and VERTICAL COMPONENTS. Having these COMPONENTS defined will allow for us to find the overall MAGNITUDE of the VELOCITY, which in this problem, is asked to be defined at 1.25 seconds after launch. To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine VELOCITY of the PROJECTILE at any point in its TRAJECTORY, they are: 𝑣L = 𝑣8 cos 𝜃 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃
Where: • 𝑣8 = INITIAL VELOCITY • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT
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• 𝑡 = TIME In our problem statement we are given all of this DATA, which is: • 𝑣8 = 50 m/s • 𝜃 = 25° • 𝑔 = 9.81 m/s A • 𝑡 = 1.25 s Let’s take this information and determine each of our VELOCITY COMPONENTS, starting with our HORIZONTAL COMPONENT, which will be: 𝑣L = (50 m/s) cos 25° Or: 𝑣L = 45.3 m/s Doing the same for our VERTICAL COMPONENT, we have: 𝑣M = −(9.81 m/s A )(1.25 s) + (50 m/s) sin 25° Or: 𝑣M = 8.87 m/s
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Its important to NOTE the CHANGE in VELOCITY relative to the INITIAL VELOCITY at LAUNCH. What this is telling us is that the VERTICAL VELOCITY at this POINT in the TRAJECTORY has slowed significantly. Although it is still traveling in the same POSITIVE DIRECTION, VERTICALLY, it will continue to slow until it reaches a VELOCITY of ZERO and begin it’s decent in the OPPOSITE direction…DOWNWARDS. You may also notice that there was no TIME VARIABLE tied to the HORIZONTAL COMPONENT of VELOCITY. Let’s recall that when working PROJECTILE MOTION problems, we are analyzing under IDEAL CONDITIONS and the assumption that AIR RESISTANCE is NEGLIGIBLE. The only FORCE of significance that acts on any object, or PARTICLE for that matter, is GRAVITY. Gravity will always act VERTICALLY DOWNWARD from the center of the OBJECTS MASS causing a DOWNWARD ACCELERATION back towards the surface of the EARTH, or the realm in which the PARTICLE resides. Due to the objects INERTIA created by the initial projection, and the fact that we are working under the assumption that AIR RESISTANCE is NEGLIGABLE, there is created an instance where there is no need for an external horizontal force to maintain the HORIZONTAL COMPONENT of the objects VELOCITY…it remains constant.
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However, the VERTICAL COMPONENT of objects VELOCITY is under CONSTANT GRAVITATIONAL ACCELERATION meaning that it will reduce over TIME. At this POINT, we now have defined the HORIZONTAL and VERTICAL COMPONENTS of VELOCITY 1.25 seconds after LAUNCH, which are: 𝑣L = 45.3 m/s 𝑣M = 8.87 m/s Solving for the MAGNITUDE of the VELOCITY, we can use the FORMULA:
𝑣=
𝑣L A + 𝑣M A
Which gives us:
𝑣=
(45.3 m/s)A + (8.87 m/s)A
Or: 𝑣 = 46.2 m/s The velocity of this projectile after 1.25 seconds in flight is 46.2 m/s. The correct answer choice is C. 46.2 m/s
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PROBLEM 1: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the range of the projectile is closest to: (m)
A. 87 B. 160 C. 214 D. 471
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PROBLEM 2: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the highest point of the trajectory is closest to:
A. 105 m B. 195 m C. 305 m D. 655 m
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PROBLEM 3: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the velocity of this projectile after 2 seconds in flight is closest to:
A. 8.2 m/s B. 19.6 m/s C. 29.7 m/s D. 53 m/s
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PROBLEM 4: A cannon is launched from a hillside with an elevation of 143 m in route to a target with an elevation of 43 m. If the cannon is launched with a velocity of 30 m/s at an angle of 20o above the horizontal, the angle in which this projectile makes with the horizontal axis after 2 seconds in flight is closest to:
A. 188.4° B. 94.1° C. 7.7° D. 18.4°
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PROBLEM 5: A football is thrown at the line of scrimmage with a velocity of 22 m/s at an angle of 26o above the horizontal toward a receiver standing still 35 yards downfield. If the quarterback launches the ball from a point 8 ft above the ground, the height at which the receiver must catch this ball is closest to: A. 5.9 ft B. 8.7 ft C. 2.65 ft D. 13 ft
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PROBLEM 6: A ball is thrown with a velocity of 7 m/s toward a wall that is 5 meters away. If the ball is propelled from a point that is 3 meters above the ground at an angle of 34o above the horizontal, the height at which the ball will hit the wall is closest to:
A. 2.73 m B. 1.44 m C. 0.27 m D. 0.11 m
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PROBLEM 7: A projectile is launched from a cliff 200 meters above level ground with a launch velocity of 50 m/s and a launch angle of 25o above the horizontal. Assuming negligible air resistance, the range of this projectile before making it impact is closest to: (m)
A. 517 B. 51.7 C. 1,517 D. 767
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PROBLEM 8: A projectile is launched from a cliff 200 meters above level ground with a launch velocity of 50 m/s and a launch angle of 25o above the horizontal. Assuming negligible air resistance, the maximum height of its trajectory is closest to: (m)
A. 223 B. 157 C. 336 D. 761
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PROBLEM 9: A projectile is launched from a cliff 200 meters above level ground with a launch velocity of 50 m/s and a launch angle of 25o above the horizontal. Assuming negligible air resistance, the velocity after 1.25 seconds after launch is closest to: (m/s)
A. 8.1 B. 16.3 C. 46.2 D. 102
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PROJECTILE MOTION | SOLUTIONS
SOLUTION 1: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away.
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We know through our studies in PROJECTILE MOTION, that this projectile will follow some defined TRAJECTORY before impact, such that:
We are being asked to determine the RANGE of this projectile before making an impact on the intended target. Or in other words, HOW FAR will this projectile TRAVEL HORIZONTALLY before smashing its target. This may seem like a complicated endeavor that we are about to embark on, but luckily, the process is clean and easily replicated across a number of scenarios, let’s hash it out. We know right off the bat that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT.
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If we flip over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement, although it may not be readily obvious, its there. We can also set the POSITION of the LAUNCH POINT at any position that is convenient for us, as long as we account for that precise location in our computations moving forward.
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Let’s set our LAUNCH POINT to be located at: (𝑥8 , 𝑦8 ) = (0,0) So we are saying that our cannonball will be launched at the COORDINATES of (0,0). All measurements will be relative to this POINT, as identified by the AXIS we have established:
This gives us the ability to DEFINE the relative DATA right up front, which will be: • 𝑣8 = 30 m/s • 𝑥8 = 0 • 𝑦8 = 0 • 𝜃 = 20° • 𝑔 = 9.81 m/s A
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We can also conclude that based on the information that we are presented, that the VERTICAL DISPLACEMENT, y, at the end of the TRAJECTORY will be: 𝑦 = 43 m − 143 m Or: 𝑦 = −100 m It’s important to NOTE the NEGATIVE SIGN here, the ELEVATION DROPS, therefore, there will be a NEGATIVE DISPLACEMENT experienced VERTICALLY by this PROJECTILE. If this is not picked up, our calculations from the start will be incorrect, affecting everything that comes down the line. With this VERTICAL DISPLACEMENT defined for us, let’s revisit our GENERAL FORMULA given to us in the NCEES Reference Handbook, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Which we have defined: • 𝑣8 = 30 m/s • 𝑦8 = 0 • 𝜃 = 20°
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• 𝑔 = 9.81 m/s A • 𝑦 = −100 m We have everything we need with the exception of the TIME, t, so let’s plug and chug some numbers to quantify this VALUE. Plugging in our data we have: (9.81 m/s A )𝑡 A −100 m = − + (30 m/s) sin 20° 𝑡 + 0 2 Simplifying this expression, and removing the UNITS for the time being, we get: −100 m = −4.91𝑡 A + 10.26𝑡 We now have a QUADRATIC EQUATION that we can solve using the QUADRATIC FORMULA, which in the context of our VARIABLE for TIME, t, is written as: −𝑏 ± 𝑏 A − 4𝑎𝑐 𝑡= 2𝑎 Where: 𝑎 = −4.91 𝑏 = 10.26 𝑐 = 100
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Plugging these VALUES in to QUADRATIC FORMULA, we have: −(10.26) ± (10.26)A − 4(−4.91)(100) 𝑡= 2(−4.91) Which gives us the TWO SOLUTIONS: 𝑡I = −3.59 s 𝑡A = 5.68 s We obviously can’t go “back” in TIME, so the negative solution is irrelevant, telling us that the PROJECTILE is in MOTION for 5.68 SECONDS until it makes an IMPACT at the location of the TARGET. We can now refer back to our GENERAL FORMULA given to us in the NCEES Reference Handbook for the HORIZONTAL DISPLACEMENT, which is: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 Where: • 𝑣8 = 30 m/s • 𝑥8 = 0 • 𝜃 = 20° • 𝑡 = 5.68 s
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We have all the necessary DATA defined and are now at a point of a simple plug and chug of our numbers to quantify the RANGE. Plugging in our data we have: 𝑥 = 30 m/s cos 20° (5.68 s) + 0 Or that: 𝑥 = 160 m The RANGE of the projectile before impact at the location of the target is 160 m. The correct answer choice is B. 160 m
SOLUTION 2: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant.
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In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away.
We know through our studies in PROJECTILE MOTION, that the cannonball will follow a defined TRAJECTORY, and at some defined point, 𝑦JKL , it will be at its PEAK, such that it changes from a direction of traveling upwards to one that is traveling downwards. This PEAK height above the ELEVATION of the TARGET is what we are after:
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To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine when this PEAK occurs and the MEASUREMENT at this POINT, they are: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Our approach will be to take the INITIAL VELOCITY at the LAUNCH POINT, along with its associated GEOMETRY, break it down it to its VERTICAL and HORIZONTAL COMPONENTS, and determine where the VERTICAL VELOCITY is ZERO. By observation, we know that at IMPACT, that this will be the case…but it will also be the case at the TOP of the TRAJECTORY, 𝑦JKL , where the VELOCITY in the VERTICAL DIRECTION transitions from going UPWARD to DOWNWARD. There is an INSTANTANEOUS VELOCITY of ZERO at this particular point, and the TIME it takes to get to that POINT is what we are initially after, so let’s get to work.
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Taking the full TRAJECTORY of the CANNON, let’s hone in on the LAUNCH POINT, such that:
Breaking down the INITIAL VELOCITY, we find that the HORIZONTAL and VERTICAL COMPONENTS will be: 𝑣8L = 𝑣8 cos 𝜃 = (30 m/s) cos 20° 𝑣8M = 𝑣8 sin 𝜃 = (30 m/s) sin 20° Or: 𝑣8L = 28.2 m/s 𝑣8M = 10.26 m/s Now we aren’t concerned with the HORIZONTAL COMPONENT at this point, but we do want to dive in to analyzing how that VERTICAL COMPONENT will be affected.
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As we have established from the NCEES Reference Handbook, the HORIZONTAL COMPONENT of VELOCITY at ANY POINT in the TRAJECTORY can be expressed as: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 Where: 𝑣8 sin 𝜃 = 𝑣8M So: 𝑣M = −𝑔𝑡 + 𝑣8M We know that: 𝑣8M = 10.26 m/s 𝑔 = 9.81 m/s A And that at the PEAK of TRAJECTORY: 𝑣M = 0 m/s Plugging all of these values in to the FORMULA, we get: 0 = −(9.81 m/s A )𝑡 + 10.26 m/s
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Rearranging to isolate and solve for the TIME to this POINT, we get:
𝑡=
10.26 m/s = 1.05 s 9.81 m/s A
Knowing this, we can now refer back to the second formula we pulled from the NCEES Reference Handbook, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Where: • 𝑣8 = 30 m/s • 𝑦8 = 100 m • 𝜃 = 20° • 𝑔 = 9.81 m/s A • 𝑡 = 1.05 s Simply plugging in this DATA, we have: (9.81 m/s A )(1.05 s)A 𝑦=− + (30 m/s) sin 20° (1.05 s) + 100 m 2 Which gives us: 𝑦 = 105 m
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The HIGHEST POINT of the trajectory will be 105 meters above the ELEVATION of the TARGET. The correct answer choice is A. 105 m
SOLUTION 3: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away:
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We know through our studies in PROJECTILE MOTION, that the cannonball will follow a defined TRAJECTORY and at any point in that TRAJECTORY it will have a VELOCITY that can be broken up in to its individual HORIZONTAL and VERTICAL COMPONENTS. Having these COMPONENTS defined will allow for us to find the overall MAGNITUDE of the VELOCITY, which in this problem, is asked to be defined 2 seconds after launch:
To start, let’s flip over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine VELOCITY of the PROJECTILE at any point in its TRAJECTORY, they are: 𝑣L = 𝑣8 cos 𝜃 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃
Where: • 𝑣8 = INITIAL VELOCITY
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• 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME In our problem statement we are given all of this DATA, which is: • 𝑣8 = 30 m/s • 𝜃 = 20° • 𝑔 = 9.81 m/s A • 𝑡 = 2 s Let’s take this information and determine each of our VELOCITY COMPONENTS, starting with our HORIZONTAL COMPONENT, which will be: 𝑣L = (30 m/s) cos 20° Or: 𝑣L = 28.2 m/s Doing the same for our VERTICAL COMPONENT, we have: 𝑣M = −(9.81 m/s A )(2 s) + (30 m/s) sin 20°
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Or: 𝑣M = −9.36 m/s Its important to NOTE here the NEGATIVE SIGN in front of our VERTICAL VELOCITY at this POINT. What this is telling us is that the VERTICAL VELOCITY at this POINT in the TRAJECTORY is going in the OPPOSITE DIRECTION then what we had established as the POSITIVE DIRECTION at the beginning of our problem. In other words, we established that the INITIAL VELOCITY going UPWARDS was POSITIVE…the NEGATIVE sign is telling us that now the VELOCITY of the projectile has it headed DOWNWARDS…for this COMPONENT. You may also notice that there was no TIME VARIABLE tied to the HORIZONTAL COMPONENT of VELOCITY. Let’s recall that when working PROJECTILE MOTION problems, we are analyzing under IDEAL CONDITIONS and the assumption that AIR RESISTANCE is NEGLIGIBLE. The only FORCE of significance that acts on any object, or PARTICLE for that matter, is GRAVITY.
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Gravity will always act VERTICALLY DOWNWARD from the center of the OBJECTS MASS causing a DOWNWARD ACCELERATION back towards the surface of the EARTH, or the realm in which the PARTICLE resides. Due to the objects INERTIA created by the initial projection, and the fact that we are working under the assumption that AIR RESISTANCE is NEGLIGABLE, there is created an instance where there is no need for an external horizontal force to maintain the HORIZONTAL COMPONENT of the objects VELOCITY…it remains CONSTANT. However, the VERTICAL COMPONENT of the objects VELOCITY is under CONSTANT GRAVITATIONAL ACCELERATION meaning that it will reduce over TIME. At this POINT, we now have defined the HORIZONTAL and VERTICAL COMPONENTS of VELOCITY 2 seconds after LAUNCH, which are: 𝑣L = 28.2 m/s 𝑣M = −9.36 m/s Solving for the MAGNITUDE of the VELOCITY, we can use the FORMULA:
𝑣=
𝑣L A + 𝑣M A
Which gives us:
𝑣=
(28.2 m/s)A + (−9.36 m/s)A
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Or: 𝑣 = 29.7 m/s The velocity of this projectile after 2 seconds in flight is 29.7 m/s. The correct answer choice is C. 29.7 m/s
SOLUTION 4: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from a cannon located at some elevation, horizontally, towards a target that is some unknown distance away.
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We know through our studies in PROJECTILE MOTION, that the cannonball will follow a defined TRAJECTORY and at any point in that TRAJECTORY it will have a VELOCITY that can be broken up in to its individual HORIZONTAL and VERTICAL COMPONENTS. Having these COMPONENTS defined will allow for us to find the ANGLE in which the VELOCITY VECTOR is making with the HORIZONTAL AXIS 2 seconds after launch:
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To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine VELOCITY of the PROJECTILE at any point in its TRAJECTORY, they are: 𝑣L = 𝑣8 cos 𝜃 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃
Where: • 𝑣8 = INITIAL VELOCITY • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME In our problem statement we are given all of this DATA, which is: • 𝑣8 = 30 m/s • 𝜃 = 20° • 𝑔 = 9.81 m/s A • 𝑡 = 2 s Let’s take this information and determine each of our VELOCITY COMPONENTS, starting with our HORIZONTAL COMPONENT, which will be: 𝑣L = (30 m/s) cos 20°
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Or: 𝑣L = 28.2 m/s Doing the same for our VERTICAL COMPONENT, we have: 𝑣M = −(9.81 m/s A )(2 s) + (30 m/s) sin 20° Or: 𝑣M = −9.36 m/s Its important to NOTE here the NEGATIVE SIGN in front of our VERTICAL VELOCITY at this POINT. What this is telling us is that the VERTICAL VELOCITY at this POINT in the TRAJECTORY is going in the OPPOSITE DIRECTION then what we had established as the POSITIVE DIRECTION at the beginning of our problem. In other words, we established that the INITIAL VELOCITY going UPWARDS was POSITIVE…the NEGATIVE sign is telling us that now the VELOCITY is headed DOWNWARDS…for this COMPONENT. You may also notice that there was no TIME VARIABLE tied to the HORIZONTAL COMPONENT of VELOCITY.
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Let’s recall that when working PROJECTILE MOTION problems, we are analyzing under IDEAL CONDITIONS and the assumption that AIR RESISTANCE is NEGLIGIBLE. The only FORCE of significance that acts on any object, or PARTICLE for that matter, is GRAVITY. Gravity will always act VERTICALLY DOWNWARD from the center of the OBJECTS MASS causing a DOWNWARD ACCELERATION back towards the surface of the EARTH, or the realm in which the PARTICLE resides. Due to the objects INERTIA created by the initial projection, and the fact that we are working under the assumption that AIR RESISTANCE is NEGLIGABLE, there is created an instance where there is no need for an external horizontal force to maintain the HORIZONTAL COMPONENT of the objects VELOCITY…it remains CONSTANT. However, the VERTICAL COMPONENT of the objects VELOCITY is under CONSTANT GRAVITATIONAL ACCELERATION meaning that it will reduce over TIME. At this POINT, we now have defined the HORIZONTAL and VERTICAL COMPONENTS of VELOCITY 2 seconds after LAUNCH, which are: 𝑣L = 28.2 m/s 𝑣M = −9.36 m/s
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Referring back to our TRIGONOMETRY RELATIONSHIPS, we know that:
tan(𝜃) =
𝑦 𝑥
Or in terms of the famous adage SOH CAH TOA:
tan(𝜃) =
opposite adjacent
Whichever way you remember best is the one we want you to roll with. Taking our COMPONENTS, we can illustrate the GEOMETRY of the VELOCITY VECTOR at this POINT as:
At this POINT in the TRAJECTORY the projectile is going to the RIGHT and DOWNWARD, meaning that the ANGLE with the HORIZONTAL AXIS will be BELOW the AXIS, as illustrated.
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Our COMPONENTS can then be defined as: Adjacent = 𝑣L = 28.2 m/s Opposite = 𝑣M = 9.36 m/s Plugging these into our TRIGONOMETRIC RELATIONSHIP for the TANGENT, we have:
tan(𝜃) =
9.36 m/s 28.2 m/s
Rearranging to ISOLATE and SOLVE for the ANGLE, we get:
𝜃 = tanUI
9.36 m/s 28.2 m/s
Or: 𝜃 = 18.4° Now the important characteristic to NOTE here is the LOCATION of this ANGLE. Come exam day, it may be measured starting at the POSITIVE X-AXIS and COUNTERCLOCKWISE, making it 341.6°. It may also be presented as an answer option with a NEGATIVE SIGN, indicating its LOCATION relative to the POSITIVE X-AXIS.
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Nonetheless, the ANGLE that the VELOCITY VECTOR of this projectile makes with the HORIZONTAL AXIS after 2 seconds in flight is 18.4°. The correct answer choice is D. 18.40
SOLUTION 5: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario where a quarterback is throwing a football 35 yards downfield to a receiver standing still. Given the information presented to us in the problem statement, we want to know at what elevation the receiver must go to catch this ball at the point he is located. This may seem like a complicated task, but luckily, the process is clean and easily replicated across a number of scenarios, let’s hash it out.
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We know right off the bat that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT. Taking a look at the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement, although it may not be directly apparent, its there.
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We can also set the POSITION of the AXIS in which the LAUNCH POINT is measured to any point that is convenient for us, as long as we account for that precise location in our computations moving forward. Let’s set our LAUNCH AXIS at the feet of the quarterback, and directly below the HORIZONTAL position of the LAUNCH POINT. This makes our COORDINATES then: (𝑥8 , 𝑦8 ) = ( 0, 8 ft) All measurements will be relative to this AXIS. We can now DEFINE the relative DATA presented to us, which is: • 𝑣8 = 22 m/s • 𝑥8 = 0 • 𝑦8 = 8 ft • 𝜃 = 26° • 𝑔 = 9.81 m/s A We also know that: • 𝑥 = 35 yards Various UNITS of measure are being used here, so let’s convert everything in to SI to make sure all calculations moving forward are accurate.
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This will be a point that many students will muddy up a fairly straight forward problem, so keep calm under the pressure and go through your mental checklists to ensure everything is laying out correctly. Converting our HORIZONTAL DISPLACEMENT as well as the VERTICAL LAUNCH POINT, we get: • 𝑥 = 38.3 m • 𝑦8 = 2.43 m The most common CONVERSION FACTORS can be found on PAGE 2 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Everything is now looking good, let’s revisit the formulas we pulled from the PARTICLE KINEMATICS section once again, they are: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
We have most everything we need to just revert to a plug and play problem.
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With the DISPLACEMENT formulas presented to us, let’s create a pair of expressions representing our scenario, this gives us: 38.3 m = (22 m/s) cos 26° 𝑡 + 0 (9.81 m/s A )𝑡 A 𝑦=− + (22 m/s) sin 26° 𝑡 + 2.43 m 2
We are now at a point where we have TWO EQUATIONS and TWO UNKNOWNS…everything is set up perfectly for us to quantify these two values, of which, we are ultimately concerned with defining the HEIGHT in which the receiver must go to catch this ball. With this pair of EQUATIONS set up, we would recommend from here that you learn how to HACK this problem, solving it in a FRACTION of the time using your NCEES APPROVED CALCULATOR. As a student of PREPINEER, you have full access to seeing how this is done on your calculator of choice using the CALCULATOR HACKS that we have developed and archived for your ON-DEMAND TRAINING. But we will still walk through the step by step calculations because we believe in the importance of fully understanding the fundamentals behind all the work that we do, first and foremost, so let’s do just that.
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Simplifying the expressions that we have derived and because we like to work with just numbers, let’s keep in mind the UNITS we are working with, but drop them for now: 38.3 = 19.8𝑡 𝑦 = −4.91𝑡 A + 9.64𝑡
Alright, so those look a lot friendlier to the eye. With only a SINGLE UNKNOWN VARIABLE, let take the FIRST EXPRESSION and rearrange it to ISOLATE and solve for TIME, t. Doing so we get: 𝑡 = 1.94 s Taking this newly defined value for TIME and plugging it in to the SECOND EXPRESSION, we get: 𝑦 = −4.91 1.94 s
A
+ 9.64 1.94 s + 2.43 m
Or that: 𝑦 = 2.65 m So given that a quarterback chucks a football from the line of scrimmage with a velocity of 22 m/s and at an angle of 26o above the horizontal, it must be received at a point that
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is 2.65 meters above the ground…or since the answer options are given in feet, converting back to this UNIT, the reception must be made at a point that is 8.7 feet above the ground. The correct answer choice is B. 8.7 ft
SOLUTION 6: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario where a ball is thrown at a wall that is 5 meters away. Given the information presented to us in the problem statement, we want to know at what elevation the ball will make impact with the wall. By now, it’s clear that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT.
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Turning to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement. We can set the POSITION of the AXIS in which the LAUNCH POINT is measured to any point that is convenient for us, as long as we account for that precise location in our computations moving forward. Let’s set our AXIS at the actual LAUNCH POINT.
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This makes our COORDINATES then: (𝑥8 , 𝑦8 ) = ( 0, 0) All measurements will be relative to this POINT, as identified by the AXIS we have established. This gives us the ability to DEFINE the relative DATA right up front, which will be: • 𝑣8 = 7 m/s • 𝑥8 = 0 • 𝑦8 = 0 • 𝜃 = 34° • 𝑔 = 9.81 m/s A We also know that: • 𝑥 = 5 meters Everything is looking good at this point, so let’s revisit the formulas we pulled from the PARTICLE KINEMATICS section once again, which are: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
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We have most everything we need to just revert to a plug and play problem. With the DISPLACEMENT formulas presented to us, let’s create a pair of expressions representing this scenario, giving us: 5 m = (7 m/s) cos 34° 𝑡 + 0 (9.81 m/s A )𝑡 A 𝑦=− + (7 m/s) sin 34° 𝑡 + 0 2
We are now at a point where we have TWO EQUATIONS and TWO UNKNOWNS…everything is set up perfectly for us to quantify these two values, of which, we are ultimately concerned with defining the HEIGHT in which the ball strikes the wall. With this pair of EQUATIONS set up, we would recommend from here that you learn how to HACK this problem, solving it in a FRACTION of the time using your NCEES APPROVED CALCULATOR. As a student of PREPINEER, you have full access to seeing how this is done on your calculator of choice using the CALCULATOR HACKS that we have developed and archived for your ON-DEMAND TRAINING. But we will still walk through the step by step calculations because we believe in the importance of fully understanding the fundamentals behind all the work that we do, first and foremost, so let’s do just that.
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Simplifying the expressions that we have derived and because we like to work with just numbers, let’s keep in mind the UNITS we are working with, but drop them for now: 5 = 5.8𝑡 𝑦 = −4.91𝑡 A + 3.91𝑡
With only a SINGLE UNKNOWN, take the FIRST EXPRESSION and rearrange it to ISOLATE and solve for the VARIABLE t.
Doing so we get:
𝑡 = .86 s Taking this value for TIME and plugging it in to the SECOND EXPRESSION, we get: 𝑦 = −4.91 . 86
A
+ 3.91(.86)
Or that: 𝑦 = −.27 m Now this answer might surprise you, and certainly would make you doubt the work that you have done up to this point.
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Many will stop, reconsider doing all their calculations over, and go from there…some will just quit and go to the next problem. However, all is well here, all we need to do is realize that the ball is being released at a POINT that is 3 meters above the ground. Our calculations are telling us that the height of the ball will be .27 meters below this starting point, or that: 𝑦 = 3 m − .27 m = 2.73 m The ball will strike at a point that is 2.73 meters above the ground. The correct answer choice is A. 2.73 m
SOLUTION 7: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant.
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In this problem, we are given an illustration and scenario of a projectile being launched from some contraption located 200 meters above the ground, horizontally, towards a target that is some unknown distance away. We are being asked to determine the RANGE of this projectile before making an impact. Or in other words, HOW FAR this projectile will TRAVEL HORIZONTALLY before smashing in to the ground. We inherently know that this PROJECTILE will experience some HORIZONTAL and VERTICAL DISPLACEMENT. Flipping back to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we can pull a few GENERAL FORMULAS that will allow us to determine what these DISTANCES, or DISPLACEMENTS, will be: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Where: • 𝑣8 = INITIAL VELOCITY • 𝑥8 = HORIZONTAL POSITION at the LAUNCH POINT • 𝑦8 = VERTICAL POSITION at the LAUNCH POINT Made with by Prepineer | Prepineer.com
• 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT • 𝑡 = TIME We have a lot of this DATA already defined for us in the problem statement. We can set the POSITION of the LAUNCH POINT at any point that is convenient for us, as long as we account for that precise location in our computations moving forward. Let’s set our AXIS, relative to our LAUNCH POINT, at the location: (𝑥8 , 𝑦8 ) = ( 0, 200 ) Essentially this would be at the ground level, directly below the LAUNCH POINT…indicating that the PROJECTILE has a starting POSITION that is ZERO UNITS in the HORIZONTAL DIRECTION and 200 UNITS in the VERTICAL DIRECTION. All measurements will be relative to this AXIS. This gives us the ability to DEFINE the relative DATA right up front, which will be: • 𝑣8 = 50 m/s • 𝑥8 = 0 • 𝑦8 = 200 m • 𝜃 = 25° • 𝑔 = 9.81 m/s A
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With this information defined, let’s revisit our GENERAL FORMULA given to us in the NCEES Reference Handbook for the HORIZONTAL DISPLACEMENT, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Of the variables needed, we have: • 𝑣8 = 50 m/s • 𝑦8 = 200 m • 𝜃 = 25° • 𝑔 = 9.81 m/s A We also know that this PROJECTILE will come crashing down at an ELEVATION that is 200 meters below where it started, therefore: 𝑦 = −200 m We have everything we need with the exception of the TIME, t, so let’s plug and chug some numbers to quantify this VALUE. Plugging in our data we have: (9.81 m/s A )𝑡 A −200 m = − + (50 m/s) sin 25° 𝑡 + 200 m 2
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Simplifying this expression, and removing the UNITS for the time being, we get: −400 m = −4.91𝑡 A + 21.1𝑡 We now have a QUADRATIC EQUATION that we can solve using the QUADRATIC FORMULA, which in the context of our VARIABLE for TIME, t, is written as: −𝑏 ± 𝑏 A − 4𝑎𝑐 𝑡= 2𝑎 Where: 𝑎 = −4.91 𝑏 = 21.1 𝑐 = 400 Plugging these VALUES in to QUADRATIC FORMULA, we have: −(21.1) ± (21.1)A − 4(−4.91)(400) 𝑡= 2(−4.91) Which gives us the TWO SOLUTIONS: 𝑡I = −7.13 s 𝑡A = 11.4 s
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We obviously can’t go “back” in TIME, so the negative solution is irrelevant, telling us that the PROJECTILE is in MOTION for 11.4 SECONDS until it makes an IMPACT at the location of the TARGET. We can now refer back to our GENERAL FORMULA given to us in the NCEES Reference Handbook for the HORIZONTAL DISPLACEMENT, which is: 𝑥 = 𝑣8 cos 𝜃 𝑡 + 𝑥8 Which we have defined: • 𝑣8 = 50 m/s • 𝑥8 = 0 • 𝜃 = 25° • 𝑡 = 11.4 s Plugging in our numbers to quantify the RANGE, we have: 𝑥 = (50 m/s) cos 25° (11.4 s) + 0 Or that: 𝑥 = 517 m The RANGE of the projectile before impact at the location of the target is 517 m. The correct answer choice is A. 517 m
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SOLUTION 8: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant. In this problem, we are given an illustration and scenario of a projectile being launched from some contraption located 200 meters above the ground, horizontally, towards a target that is some unknown distance away. We know through our studies in PROJECTILE MOTION, that the projectile will follow a defined TRAJECTORY, and at some defined point, 𝑦JKL , it will be at its PEAK, such that it changes from a direction of traveling UPWARDS to one that is traveling DOWNWARDS. This PEAK height above the ELEVATION of the TARGET is what we are after.
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To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine when this PEAK occurs and the MEASUREMENT at this POINT, they are: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2
Our approach will be to take the INITIAL VELOCITY at the LAUNCH POINT, along with its associated GEOMETRY, break it down it to its VERTICAL and HORIZONTAL COMPONENTS, and determine where the VERTICAL VELOCITY is ZERO. By observation, we know that at IMPACT, that this will be the case…but it will also be the case at the TOP of the TRAJECTORY, 𝑦JKL , where the VELOCITY in the VERTICAL DIRECTION transitions from going UPWARD to DOWNWARD. There is an INSTANTANEOUS VELOCITY of ZERO at this particular point, and the TIME it takes to get to that POINT is what we are initially after, so let’s get to work. Breaking down the INITIAL VELOCITY, we find that the HORIZONTAL and VERTICAL COMPONENTS will be: 𝑣8L = 𝑣8 cos 𝜃 = (50 m/s) cos 25° 𝑣8M = 𝑣8 sin 𝜃 = (50 m/s) sin 25°
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Or: 𝑣8L = 45.3 m/s 𝑣8M = 21.1 m/s Now we aren’t concerned with the HORIZONTAL COMPONENT at this point, but we do want to dive in to analyzing how that VERTICAL COMPONENT will be affected. As we have established from the NCEES Reference Handbook, the HORIZONTAL COMPONENT of VELOCITY at ANY POINT in the TRAJECTORY can be expressed as: 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃 Where: 𝑣8 sin 𝜃 = 𝑣8M So: 𝑣M = −𝑔𝑡 + 𝑣8M We know that: 𝑣8M = 21.1 m/s 𝑔 = 9.81 m/s A
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And that at the PEAK of TRAJECTORY: 𝑣M = 0 m/s Plugging all of these values in to the FORMULA, we get: 0 = −(9.81 m/s A )𝑡 + 21.1 m/s Rearranging to isolate and solve for the TIME to this POINT, we get:
𝑡=
21.1 m/s = 2.15 s 9.81 m/s A
Knowing this, we can now refer back to the second formula we pulled from the NCEES Reference Handbook, which is: 𝑔𝑡 A 𝑦=− + 𝑣8 sin 𝜃 𝑡 + 𝑦8 2 Where we have defined: • 𝑣8 = 50 m/s • 𝑦8 = 200 m • 𝜃 = 25° • 𝑔 = 9.81 m/s A • 𝑡 = 2.15 s
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Simply plugging in this DATA, we have: (9.81 m/s A )(2.15 s)A 𝑦=− + (50 m/s) sin 25° (2.15 s) + 200 m 2 Which gives us: 𝑦 = 223 m The HIGHEST POINT of the trajectory is 223 meters above the ELEVATION of the TARGET. The correct answer choice is A. 223 m
SOLUTION 9: The TOPIC of PROJECTILE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. PROJECTILE MOTION is a form of CURVILINEAR TRANSLATIONAL MOTION wherein which an object, or particle, referred to as a PROJECTILE, is propelled in some direction near the Earth’s surface, or any surface with an established gravitational constant.
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In this problem, we are given an illustration and scenario of a projectile being launched from some contraption located 200 meters above the ground, horizontally, towards a target that is some unknown distance away. We know through our studies in PROJECTILE MOTION, that the projectile will follow a defined TRAJECTORY and at any point in that TRAJECTORY it will have a VELOCITY that can be broken up in to its individual HORIZONTAL and VERTICAL COMPONENTS. Having these COMPONENTS defined will allow for us to find the overall MAGNITUDE of the VELOCITY, which in this problem, is asked to be defined at 1.25 seconds after launch. To start, let’s hop over to the section titled PARTICLE KINEMATICS, on PAGE 74 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and pull a few GENERAL FORMULAS that will allow us to determine VELOCITY of the PROJECTILE at any point in its TRAJECTORY, they are: 𝑣L = 𝑣8 cos 𝜃 𝑣M = −𝑔𝑡 + 𝑣8 sin 𝜃
Where: • 𝑣8 = INITIAL VELOCITY • 𝜃 = LAUNCH ANGLE • 𝑔 = GRAVITATIONAL CONSTANT
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• 𝑡 = TIME In our problem statement we are given all of this DATA, which is: • 𝑣8 = 50 m/s • 𝜃 = 25° • 𝑔 = 9.81 m/s A • 𝑡 = 1.25 s Let’s take this information and determine each of our VELOCITY COMPONENTS, starting with our HORIZONTAL COMPONENT, which will be: 𝑣L = (50 m/s) cos 25° Or: 𝑣L = 45.3 m/s Doing the same for our VERTICAL COMPONENT, we have: 𝑣M = −(9.81 m/s A )(1.25 s) + (50 m/s) sin 25° Or: 𝑣M = 8.87 m/s
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Its important to NOTE the CHANGE in VELOCITY relative to the INITIAL VELOCITY at LAUNCH. What this is telling us is that the VERTICAL VELOCITY at this POINT in the TRAJECTORY has slowed significantly. Although it is still traveling in the same POSITIVE DIRECTION, VERTICALLY, it will continue to slow until it reaches a VELOCITY of ZERO and begin it’s decent in the OPPOSITE direction…DOWNWARDS. You may also notice that there was no TIME VARIABLE tied to the HORIZONTAL COMPONENT of VELOCITY. Let’s recall that when working PROJECTILE MOTION problems, we are analyzing under IDEAL CONDITIONS and the assumption that AIR RESISTANCE is NEGLIGIBLE. The only FORCE of significance that acts on any object, or PARTICLE for that matter, is GRAVITY. Gravity will always act VERTICALLY DOWNWARD from the center of the OBJECTS MASS causing a DOWNWARD ACCELERATION back towards the surface of the EARTH, or the realm in which the PARTICLE resides. Due to the objects INERTIA created by the initial projection, and the fact that we are working under the assumption that AIR RESISTANCE is NEGLIGABLE, there is created an instance where there is no need for an external horizontal force to maintain the HORIZONTAL COMPONENT of the objects VELOCITY…it remains constant.
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However, the VERTICAL COMPONENT of objects VELOCITY is under CONSTANT GRAVITATIONAL ACCELERATION meaning that it will reduce over TIME. At this POINT, we now have defined the HORIZONTAL and VERTICAL COMPONENTS of VELOCITY 1.25 seconds after LAUNCH, which are: 𝑣L = 45.3 m/s 𝑣M = 8.87 m/s Solving for the MAGNITUDE of the VELOCITY, we can use the FORMULA:
𝑣=
𝑣L A + 𝑣M A
Which gives us:
𝑣=
(45.3 m/s)A + (8.87 m/s)A
Or: 𝑣 = 46.2 m/s The velocity of this projectile after 1.25 seconds in flight is 46.2 m/s. The correct answer choice is C. 46.2 m/s
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