Kinematics Practice Problems for AP Physics B & C

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By — McGraw-Hill Professional
Updated on Feb 12, 2011

Review the following concepts if necessary:


Multiple Choice:

  1. A firework is shot straight up in the air with an initial speed of 50 m/s. What is the maximum height it reaches?
    1. 12.5 m
    2. 25 m
    3. 125 m
    4. 250 m
    5. 1250 m
  2. 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?
    1. 2 m/s2
    2. 4 m/s2
    3. 6 m/s2
    4. 8 m/s2
    5. 10 m/s2
  3. This problem is for AP Physics C students only:

  4. An object moves such that its position is given by the function x(t) = 3t2 – 4t + 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?
    1. 32 m/s in the original direction of motion
    2. 16 m/s in the original direction of motion
    3. 0 m/s
    4. 16 m/s opposite the original direction of motion
    5. 32 m/s opposite the original direction of motion

Free Response

  1. 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.
  2. Kinematics Practice Problems/Solutions

    1. Determine the magnitude and direction of the vertical component of the bomb's velocity at the point of release.
    2. Determine the magnitude and direction of the horizontal component of the bomb's velocity at the point when the bomb contacts the target.
    3. Determine how much time it takes for the bomb to hit the target after it is released.
    4. At the point of release, what angle below the horizontal does the pilot have to look in order to see the target?


  1. Call "up" the positive direction, and set up a chart. We know that vf = 0 because, at its maximum height, the firework stops for an instant.
  2. Kinematics Practice Problems/Solutions

    Solve for Δx using equation ***: vf2 = v02 + 2ax). Answer is (C) 125 m, or about skyscraper height.

  3. Call "down" positive, and set up a chart:
  4. Kinematics Practice Problems/Solutions

    Plug straight into ** (Δx = v0t + 1/2at2) 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.

  5. First find the velocity function by taking the derivative of the position function: v(t) = 6t – 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).
    1. 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.
    2. 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.
    3. Let's call "up" the positive direction. We can solve this projectile motion problem by our able method.
    4. Kinematics Practice Problems/Solutions

      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 xx0 = v0t + 1/2at2, it's much easier to start with ***, vf2 = v02 + 2a (xx0), to find that vf 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 *(vf = v0 + at) to find that the bomb took 32 seconds to hit the ground.

    5. 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:
    6. Kinematics Practice Problems/Solutions

      By geometry, tan Δ = 2000 m/5400 m. The pilot has to look down at an angle of 20°.

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