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Dropping a tennis ball

  1. Sep 15, 2009 #1
    To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00m. It rebounds to a height of 2.00m. If the ball is in contact with the floor for 12.0ms,
    a) what is the magnitude of its average acceleration during that contact
    and
    b) is the average acceleration up or down?

    i used the equation x-x0 = v0t + 1/2 at^2. to find the time when the tennis ball just touches the ground, i did 4 = 4.9t^2, assuming down in positive and up is negative. so t = .904. then i used the equation v = v0 + at to find the velocity when the ball just touches the ground. i did v = (9.8)(.904) = 8.85m/s. the ball was in contact with the floor for 12.0ms so the final time is .904 + .012 = .916. so i have my final and initial times and i get .012ms when i subtract them. the only thing i'm missing to find the average acceleration is the final velocity. i tried using v = v0 +at again, v = 8.85 + (9.8)(.916) = 17.826m/s. so i get that the average acceleration is 748.06 upwards. but the answer in my book says 1260. how did they get that answer?
     
  2. jcsd
  3. Sep 15, 2009 #2

    rl.bhat

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    Using the formula
    v1^2 = vo^2 + 2g*h, find the velocity with which the ball hits the floor.
    Similarly to reach a height of 2 m, what should be the initial velocity after rebound? The velocity of rebound cannot be greater than the velocity with which ball hits the floor.
    It the first one is v1 and the other one is v2, then the average acceleration is ratio of change in velocity to the time. While finding the change in velocity, note down the directions of the velocities.
     
    Last edited: Sep 15, 2009
  4. Sep 15, 2009 #3
    ok so for the velocity the ball hits the ground, i get v^2 = (2)(-9.8)(-4) assuming that up is positive and down is negative so v when the ball hits the ground is 8.86m/s. now to find the initial velocity when its rebounds, i did 0 = v0^2 + (2)(-9.8)(2) because when the ball reaches 2m, that is the maximum height of that path to the velocity at 2m = 0. so i solved and i got the initial velocity when it rebounds is 6.26m/s. however when i subtract the 2 and divide by .012 seconds, i get only 216m/s^2 as opposed to what the book got which was 1260m/s^2. how is the acceleration that big?
     
  5. Sep 15, 2009 #4

    rl.bhat

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    8.86 m/s and 6.26 m/s are in the opposite direction. So the change in the velocity is
    vf - (-vi)
     
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