Review the following concepts if necessary:

- Velocity, Acceleration, and Displacement for AP Physics B & C
- Freefall for AP Physics B & C
- Projectile Motion for AP Physics B & C
- Motion Graphs for AP Physics B & C
- Kinematics: Of Special Interest to Physics C Students

### Problems

### Multiple Choice:

- A firework is shot straight up in the air with an initial speed of 50 m/s. What is the maximum height it reaches?
- 12.5 m
- 25 m
- 125 m
- 250 m
- 1250 m

- On a strange, airless planet, a ball is thrown downward from a height of 17 m. The ball initially travels at 15 m/s. If the ball hits the ground in 1 s, what is this planet's gravitational acceleration?
- 2 m/s
^{2} - 4 m/s
^{2} - 6 m/s
^{2} - 8 m/s
^{2} - 10 m/s
^{2}

- 2 m/s
- An object moves such that its position is given by the function
*x*(*t*) = 3*t*^{2}– 4*t*+ 1. The units of*t*are seconds, and the units of*x*are meters. After 6 seconds, how fast and in what direction is this object moving?- 32 m/s in the original direction of motion
- 16 m/s in the original direction of motion
- 0 m/s
- 16 m/s opposite the original direction of motion
- 32 m/s opposite the original direction of motion

**This problem is for AP Physics C students only:**

### Free Response

- An airplane attempts to drop a bomb on a target. When the bomb is released, the plane is flying upward at an angle of 30° above the horizontal at a speed of 200 m/s, as shown below. At the point of release, the plane's altitude is 2.0 km. The bomb hits the target.
- Determine the magnitude and direction of the vertical component of the bomb's velocity at the point of release.
- Determine the magnitude and direction of the horizontal component of the bomb's velocity at the point when the bomb contacts the target.
- Determine how much time it takes for the bomb to hit the target after it is released.
- At the point of release, what angle below the horizontal does the pilot have to look in order to see the target?

### Solutions

- Call "up" the positive direction, and set up a chart. We know that
*v*= 0 because, at its maximum height, the firework stops for an instant._{f} - Call "down" positive, and set up a chart:
- First find the velocity function by taking the derivative of the position function:
*v*(*t*) = 6*t*– 4. Now plug in*t*= 6 to get the velocity after six seconds; you get 32 m/s. Note that this velocity is positive. Was the object originally moving in the positive direction? Plug in*t*= 0 to the velocity formula to find out … you find the initial velocity to be –4 m/s, so the object was originally moving in the negative direction, and has reversed direction after 6 seconds. The answer is (E). -
- Because the angle 30° is measured to the horizontal, the magnitude of the vertical component of the velocity vector is just (200 m/s) (sin 30°), which is 100 m/s. The direction is "up," because the plane is flying up.
- The horizontal velocity of a projectile is constant. Thus, the horizontal velocity when the bomb hits the target is the same as the horizontal velocity at release, or (200 m/s)(cos 30°) = 170 m/s, to the right.
- Let's call "up" the positive direction. We can solve this projectile motion problem by our able method.
- Start by finding how far the bomb went horizontally. Because horizontal velocity is constant, we can say distance = velocity × time. Plugging in values from the table, distance = (170 m/s)(32 s) = 5400 m. Okay, now look at a triangle:

Don't forget to convert to meters, and be careful about directions in the vertical chart.

The horizontal chart cannot be solved for time; however, the vertical chart can. Though you could use the quadratic formula or your fancy calculator to solve

*x*–*x*=_{0}*v*+ 1/2_{0}^{t}*at*^{2}, it's much easier to start with ***,*v*_{f}^{2}=*v*_{0}^{2}+ 2*a*(*x*–*x*), to find that_{0}*v*vertically is –220 m/s (this velocity must have a negative sign because the bomb is moving down when it hits the ground). Then, plug in to *(_{f}*v*=_{f}*v*+_{0}*at*) to find that the bomb took 32 seconds to hit the ground.By geometry, tan Δ = 2000 m/5400 m. The pilot has to look down at an angle of 20°.

Solve for Δ*x* using equation ***: *v _{f}*

^{2}=

*v*

_{0}^{2}+ 2

*a*(Δ

*x*). Answer is (C) 125 m, or about skyscraper height.

Plug straight into ** (Δ*x* = *v _{0}^{t}* + 1/2

*at*

^{2}) and you have the answer. This is NOT a quadratic, because this time

*t*is a known quantity. The answer is (B) 4 m/s2, less than half of Earth's gravitational field, but close to Mars's gravitational field.

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