1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    Homework Helper

    Refer the post " dropping a tennis ball " by demonelite.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook