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What's the acceleration when it contacts the floor?

  1. Jan 8, 2004 #1
    A tennis ball is dropped from 5m High and bounces back 3.2 m high. It stays in contact with the floor for 0.036 seconds. What's the acceleration when it contacts the floor????

    The answer is 495 m/s_². I just don't know how to get to it. thx so much.
  2. jcsd
  3. Jan 8, 2004 #2
    Let [itex]v_1[/itex] be the speed of the ball just before it hits the ground, and [itex]v_2[/itex] be the speed just after.

    Using energy:

    [tex]mg(5) = \frac{1}{2} m v_1^2[/tex]
    [tex] 10g = v_1^2 [/tex]
    [tex] v_1 = \sqrt{10g} [/tex]

    [tex]mg(3.2) = \frac{1}{2} m v_2^2[/tex]
    [tex]6.4g = v_2^2 [/tex]
    [tex]v_2 = \sqrt{6.4g} [/tex]

    Acceleration is the change in velocity divided by time. (Now consider the up direction to be positive)

    [tex] a = \frac{\Delta v}{\Delta t}[/tex]
    [tex] a = \frac{\sqrt{6.4g} - (-\sqrt{10g})}{0.036}[/tex]
    [tex] a = 495 m/s^2[/tex]
  4. Jan 8, 2004 #3
    Thx so much.. I really appreciate people helping on other'S request.
    thx again

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