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Ball rebounding to less than initial height

  1. Oct 30, 2004 #1
    Hi, I have a homework problem that's giving me a bit of trouble. I hope someone can help me out and verify whether I'm correct or not :)

    After falling from rest at a height of 30 meters, a 0.50 kilogram ball rebounds upward, reaching a height of 20 meters. If the contact between ball and ground lasted 0.002 seconds, what average force was exerted on the ball?

    Ok, here is how I went about solving this. Using free-fall equation final velocity squared = initial velocity squared + (2 * gravity * height) (g=9.81), I determined that the ball impacted the ground at 24.261 m/s.

    Then, using the same equation I determined that to reach a height of 20 meters it had to leave the ground at an initial velocity 19.81 m/s.

    Thus, for average force:

    F=ma=(mass*change in velocity)/time=[0.5 kg*(19.81 m/s - (-24.261 m/s)]/0.002 seconds

    Force= 11020 Newtons

    Phew, am I correct?
     
  2. jcsd
  3. Oct 30, 2004 #2

    arildno

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    Welcome to PF!
    Looks good to me..
     
  4. Oct 30, 2004 #3

    Clausius2

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    Wow, I think you have a new medal:


    ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
    ====CONGRATULATIONS===
    ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !


    I've tried to make a Norwegian banner, but it's a bit difficult :biggrin: .
     
  5. Oct 30, 2004 #4

    arildno

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    Thank you..:shy:
    I'll start canvassing for your next one in return..
     
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