11/21-22/16Name:
Physics Test Projectile and Relative Motion—Answer Key

1. Ignoring air resistance, the horizontal component of a projectile's acceleration

A) continuously increases.

B) remains a non-zero constant.

C) is zero.

D) continuously decreases.
In projectile motion the horizontal component of the velocity is constant. Therefore, the horizontal acceleration is 0. Answer C (19/27)

2. Ignoring air resistance, the horizontal component of a projectile's velocity
A) is zero.
B) remains constant.

C) continuously increases.

D) continuously decreases.

For projectile motion, velocity in the x direction (horizontal component), or vx, is constant. Answer B (23/27)

3. A soccer ball is kicked with a velocity of 12.4 m/s at an angle of 37.8° above the horizontal. What is the vertical component of its acceleration as it travels along its trajectory?

A) (9.80 m/s2) × sin (37°) downward

B) (9.80 m/s2) × sin (37°) upward

C) 9.80 m/s2upward

D) 9.80 m/s2 downward
For projectile motion you know that motion in the x (horizontal) direction is constant and so acceleration in the x direction is always 0. In the y (vertical) direction, however, the change in velocity is related to the acceleration due to gravity. Throughout the course of the motion, no matter where in the flight the object is, the acceleration due to gravity does not change—it is always -9.8m/s2. As far as the answers go, this would be 9.80 m/s2 in the downward direction. Answer D (13/27)

4. A ball is thrown with a velocity of 20 m/s at an angle of 60° above the horizontal. What is the horizontal component of its velocity at the exact top of its trajectory?

A) zero
B) 20 m/s

C) 17 m/s

D) 10 m/s
The term instantaneous velocity refers to the velocity of the object “at the instant” the object is at the top of its flight. At that point recognize that there is only velocity in the x (horizontal) direction—velocity in the y direction at that instant is 0 m/s. So, as velocity in the horizontal direction is constant, the horizontal component of the velocity at this point will be the same as the horizontal velocity at the start of the motion, which can be determined by finding the horizontal component of the velocity. Remember that we do this by multiplying the velocity at which the projectile is launched times the cos (because the horizontal component is the adjacent side in this case) of the angle at which it was launched. Further, you should know that the cos 600 is ½ so the horizontal velocity will be ½ (20 m/s) = 10 m/s. Answer D (8/27)

5. A pilot drops a bomb from a plane flying horizontally at a constant speed. Neglecting air resistance, when the bomb hits the ground the horizontal location of the plane will

A) depend on the speed of the plane when the bomb was released.
B) be in front of the bomb.

C) be behind the bomb.

D) be over the bomb.
As the bomb is on the plane, it has a velocity in the x direction that is equal to the velocity of the plane. The instant it is released, it still has that velocity. If we ignore air resistance, the bomb will continue to travel in the x direction at the same velocity as the airplane. Therefore, when the bomb hits the ground, the plane will be directly over the bomb. Answer D (21/27)
6. A package of supplies is dropped from a plane, and one second later a second package is dropped. Neglecting air resistance, the distance between the falling packages will

A) increase.
B) decrease.
C) be constant. (only three choices)
The confusing part about this question is that you know by now that acceleration for both packages is the same—they will both accelerate at the same rate—so, the tempting thing to do would be to immediately say that as time progresses, the distance between the packages must remain constant. Trying to reason this out—simply visualize what must be happening to each package as each second passes. After 1 second, the first package has dropped a certain amount but the second package has not yet moved. After 2 seconds, the first package has dropped an amount greater than it dropped the first second, while the second package has now dropped the same amount as the first package did the first second. You should quickly realize that at each second, the second package has dropped to the point where the first package had dropped to the preceding second, and the distance between the two must be increasing. You can also draw a very quick diagram that helps you visualize this by labeling each package with subscripts to show where they are at each second (see diagram)—this only takes a few seconds and saves a lot of time trying to reason. Answer A (12/27)

7. A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?

A) 320 m

B) 160 m

C) 100 m

D) 80 m

As each package is moving at a velocity of 50.0 m/s because it is on an airplane moving at 50.0 m/s, at the instant each package is dropped, it still has a velocity in the x direction of 50.0 m/s. Each package will take the same amount of time to fall. As the second package was released two seconds after the first package, it was released (2)(50.0m) or 100 m past when the first package was released. Therefore, it will also hit 100 m past where the first package hit. Answer C (23/27)

8. A bullet is fired horizontally, and at the same instant a second bullet is dropped from the same height. Ignore air resistance. Compare the times of fall of the two bullets.
A) The fired bullet hits first.

B) The dropped bullet hits first.

C) They hit at the same time.

D) cannot tell without knowing the masses
When comparing the fall of two objects, if we ignore air resistance, both objects will be under the influence of the same acceleration due to gravity. Therefore, both objects will fall at the same rate—the motion in the x direction of bullet fired horizontally has no impact on the vertical motion of that bullet. The bullets will hit the ground at the same time. Answer C (25/27)
9. A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below?

A) It is impossible to tell from the information given.

B) Neither, since both are traveling at the same speed.
C) the stone

D) the ball
Be careful—the tendency here is to think this will be a question about the time it takes for both objects to hit the ground, which would be the same time. However, this question is asking about which object is traveling faster when it hits the ground. As the ball is dropped vertically, all it will do is accelerate at the acceleration due to gravity and its final velocity will only have this “y” component of velocity. However, the stone that is thrown from the tower will not only have the same “y” component of velocity, it will also have whatever “x” component of velocity that was added to it when it was thrown. Therefore, from the perspective of “overall” speed, the stone’s speed will be greater because it has the same y component of velocity the ball had, plus its x component of velocity. Answer C (16/27)

10. A ball is thrown at an original speed of 8.4 m/s at an angle of 32.9° above the horizontal. What is the speed of the ball when it returns to the same horizontal level?

A) 4.2 m/s

B) 8.4 m/s

C) 17 m/s

D) 9.8 m/s
We are being asked about speed here so we do not have to worry about directions being positive or negative. Also, we are not being asked about speed in the x or the y direction, just the overall speed of the ball as it reaches its original horizontal level. As we ignore air resistance, as the ball leaves the hand, while its x velocity will remain constant throughout the flight its y velocity will decrease until it reaches maximum height. Then, as the ball descends, it gain speed in the same fashion that it originally lost speed when it was ascending. As the acceleration due to gravity remains constant, when the ball reaches the same horizontal level, it should have exactly the same speed it had when it left the hand. Answer B (23/27)

11. When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed?

A) It is zero.

B) It is less than its initial speed.

C) It is equal to its initial speed.

D) It is greater than its initial speed.
We are not being asked specifically about the y or the x velocity, we are only being asked about the overall speed of the ball. As the ball leaves the foot, while its x velocity will remain constant throughout the flight, its y velocity will decrease until it reaches maximum height. At this point the y velocity will be 0 m/s. Therefore, as the overall speed is a result of the sum of the x and y velocities, the overall speed at maximum height will be less than its initial speed. Answer B (16/27)
12. If the acceleration of an object is always directed perpendicular to its velocity,

A) this situation would not be physically possible.

B) the object is speeding up.

C) the object is slowing down.

D) the object is turning.
The tendency of any object is to travel in a straight line. If a force causing an acceleration acts perpendicular to the direction of the velocity, if the acceleration is applied and then stopped, the object will then continue to move in a straight line in the slightly altered new direction after the acceleration is withdrawn. However, if the acceleration is continued, always at a direction perpendicular to the velocity, the object will be pushed into a path that is circular—so, the object is turning. The following diagram may be useful. Answer D (12/27)

13. If the acceleration vector of an object is directed anti-parallel to the velocity vector,

A) the object is turning.

B) the object is speeding up.

C) the object is slowing down.

D) the object is moving in the negative x-direction.

Anti-parallel is in a direction opposite to or at an angle of 1800 to a given vector. You are aware by now that if acceleration directly opposes the velocity vector, the velocity in the direction the object is moving will be decreasing. The positive or negative direction of the velocity is not specified in the given information so the answer will have to suffice for either. You can visualize both situations with a simple diagram. In the top diagram there is a decreasing positive velocity which is slowing down because of an opposing acceleration. In the bottom diagram there is a decreasing negative velocity which is also slowing down because of an opposing acceleration. Therefore, in both cases the result of an acceleration vector directed anti-parallel to a velocity vector is the slowing down of the object. Answer C

Quantitative Problems

In projectile motion you know that to analyze the motion you have to break the motion into x and y components and analyze each of these separately. You know that motion in the x direction is constant—that is, vx is constant. So to analyze motion in the x direction, you can only use the formula that is used for constant velocity: (although under certain circumstances you can use the range formula to find x displacement).
You know that motion in the y direction is accelerated according to the acceleration due to gravity. Therefore, to analyze motion in the y direction you can use either (from the perspective of the initial and final y velocities) or (from the perspective of displacement in the y direction).

14. A stone is thrown horizontally with an initial speed of 6.9 m/s from the edge of a cliff. A stop watch measures the stone's trajectory time from the top of the cliff to the bottom to be 5.6 s. What is the height of the cliff?

In this case, you are told that this is a horizontal launch problem, for which you immediately realize that initial y velocity is 0 m/s. You are asked to find the height of the cliff which is a y direction calculation, so you immediately know you are going to need to use: . Remember that in this case, it is not y0 that equals 0 m, it is y. Therefore, you will want to isolate y0 before you start plugging in values. Write out the y displacement formula and then isolate y0 (see below)—or, if you can write out the formula with y0 already isolated just do that. Again, remember that the vy0 term drops out. Also notice that the negative sign after isolating y0 cancels the negative sign of acceleration due to gravity.
3 points

15. A rock is thrown horizontally with a velocity of 16 m/s, from a bridge. It falls 33 m to the water below. How far does the rock travel horizontally before striking the water?

In this case, you are told that this is a horizontal launch problem, for which you immediately realize that initial y velocity is 0 m/s. You are asked to determine how far the rock travels horizontally, which is an x direction calculation, so you immediately know you are going to need to use .
NOTE-THIS IS IMPORTANT—in the vast majority of problems, if you need to
determine an x direction value (either x displacement or x velocity) you will almost always need to find time. The only way you will be able to do this is by using y direction calculations. MAKE THIS AN INTUITIVE PART OF YOUR PROBLEM SOLVING SKILLS FOR PROJECTILE MOTION. There are several possibilities:
-for horizontal launch, you will almost always need to use to find time (unless by some chance you are given the final velocity in the y direction). In this case vy0 will always be 0 m/s and the vy0tterm will drop out, and isolating t will give:

In this case, to solve for time, as this is a horizontal launch you must use .
After isolating and solving for t, plug vx and t into and solve for x.


6 points

16. A ball thrown horizontally from a point 79 m above the ground, strikes the ground after traveling horizontally a distance of 46 m. With what speed was it thrown?

In this case, you are told that this is a horizontal launch problem, for which you immediately realize that initial y velocity is 0 m/s. You are asked to determine with what velocity the rock was initially thrown, meaning what was its horizontal velocity or vx—this is an x direction calculation, so you immediately know you are going to need to use , isolating for vx. After isolating for vx you recognize that you don’t have time. Following your strategy, (see THIS IS IMPORTANT discussion for problem 15), as this is a horizontal launch problem you will need to use to find time. Isolate and solve for t, and then plug this back into your original x direction formula to solve for vx:

6 points

17. A jumper in the long-jump goes into the jump with a speed of 13.5 m/s at an angle of 23.3° above the horizontal.
(a) How long is the jumper in the air before returning to the Earth?

In this case you are told that this is an angled launch and you know you are likely to have to calculate the vx and vy0 to solve the problem, so I would normally go ahead and do this first thing so I have this information available to me (you remember without hesitation that vx = (v)(cos) and vy0 = (v)(sin). (see below for calculations).
Then, you are asked to find time. You quickly realize you would not be able to use the x direction calculation isolated for time because you do not have x displacement—you will have to use a y direction calculation. Because the launch and landing heights are the same , (see THIS IS IMPORTANT discussion for problem 15), you can use . As equals the opposite of , isolating t gives . Plug in the values for initial y velocity (which you have already calculated) and a and solve for t.

6 points (3 for vy0 and 3 for time)

(b) How far does the jumper jump?

In this case you are asked to find the x displacement, so this is an x direction calculation and you need to use vx. If you did not calculate this in part a go ahead and calculate this now. As you calculated t in the previous problem as 1.09 s, simply plug the values in and solve.


6 points (3 for vx and 3 for x)
18. A projectile is launched with an initial velocity of 45.8 m/s at an angle of 43.4° above the horizontal. What is the maximum height reached by the projectile?

In this case you are asked to find maximum height, which is a y direction calculation, and the only formula which contains y displacement is From the given information you know you can find vy0 using .
You also need to find time BUT REALIZE, you need to find time for half of the flight, not the whole flight. There are a variety of ways to do this but the easiest is to recognize that at the maximum height (which is the halfway point of the total flight), vy = 0m/s. As you already have calculated vy0, you can use , isolated for time. After find vy0 and time simply plug these into the y displacement formula and solve for maximum height.


9 points

19. A plane has an air speed of 179 m/s due North, and is in a wind of 38.6 m/s to the West. What is the plane's speed relative to the ground? (For this question, in addition to the appropriate formulas and calculations, I need to see an appropriately drawn and labeled vector diagram showing the two velocity vectors being added together and the resultant.)

We are asked to find the plane’s speed relative to the ground. The way I think about this is—the plane thinks it is traveling due north at a velocity of 179 m/s. It doesn’t realize that there is a wind blowing due west at 38.6 m/s which is changing its velocity. But, the person on the ground sees the effect of both velocities—so, we add the effect of both velocities together on a diagram to show what the person on the ground sees.