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Bouncing ball average acceleration

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A golf ball released from a height of 1.5m above a concrete floor bounces back to a height of 1.1m. If the ball is in contact with the floor for 6.2 X 10^-4 s, what is the average acceleration of the ball while in contact with the floor?

    g= 9.81 m/s^2
    x-xo= -1.5
    vo= 0
    t= 6.2 X 10^-4

    Are these variables all correct?


    2. Relevant equations
    t= ((-2(x-xo))/g)^(-1/2)
    average acceleration= (v2-v1)/(t2-t1)


    3. The attempt at a solution
    t= ((-2(-1.5m)/(9.81m/s^2))^(1/2)= 0.55s

    Do I take initial velocity divided by this time number?
    And is final velocity just 1.1 divided by time?
     
  2. jcsd
  3. Sep 14, 2009 #2

    rl.bhat

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    Homework Helper

    Using the formula
    vf^2 = vi^2 + 2gh, find vf when ball reaches the ground and vi when it rebounds from the ground.
    Then find the change in the velocity ( be careful about the directions of the velocities). Time of contact is given. Find the average acceleration.
     
  4. Sep 15, 2009 #3
    I'm still not sure what vf and vi are though. I tried dividing height by time and I used t= (-2(x-xo)/g)^1/2 to find times, but I'm still not getting the right answer.
     
  5. Sep 15, 2009 #4

    rl.bhat

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    Refer the post " dropping a tennis ball " by demonelite.
     
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