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A ball being dropped and bouncing back up?

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data
    After falling from rest at a height of 30m, a 0.50 kg ball rebounds upward, reaching a height of 20m. If the contact between ball and ground lasted 2.0 ms, what a average force was exerted on the ball?

    2. Relevant equations
    The kinematic equations?
    I dont really know

    3. The attempt at a solution
    Sorry I don't even know where to begin. I don't really understand this section in the book so i dont even know what to do. The answer is supposed to come out to 1.1 x 10^4 upward.
    Last edited: Oct 5, 2008
  2. jcsd
  3. Oct 5, 2008 #2


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    You need to work out the speed the balls hits the floor and then the speed it starts backup at. Acceleration is just change in speed / time.

    Hint - the speed it starts back up is the same as the speed a ball dropped from the second height would reach the floor.
  4. Oct 5, 2008 #3
    Ok so the speed of the ball when it hits the ground first time would need the kinematic equation Vf^2= Vi^2 + 2ad right?

    Vf^2= (0)^2 + 2(9.8)(30)
    Vf^2 = 588
    Vf~24.3 m/s ?

    Vf^2= 2(9.8)(20)
    Vf^2 = 392
    Vf ~ 19.8 m/s ?

    I think thats correct. correct me if I'm wrong please. How do I find the acceleration?
  5. Oct 5, 2008 #4


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    Acceleration = change in velocity / time
    Whats the total change in velocity (watch the signs!)
  6. Oct 5, 2008 #5

    So the acceleration is the change in velocity which would be 19.8 - 24.3/change in time.
    Then put it in the F=ma and solve for F correct? or is there more?
  7. Oct 5, 2008 #6


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    Correct method - but be careful of the signs.
  8. Oct 5, 2008 #7
    OK thank you very much. I was stuck on this problem for a while.
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