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Another use for a tennis ball

  1. Jan 23, 2008 #1
    [SOLVED] Another use for a tennis ball

    1. The problem statement, all variables and given/known data
    To test the quality of a tennis ball, you drop it onto the floor from a height of 4.93 m. It rebounds to a height of 2.68 m. If the ball is in contact with the floor for 12.3 ms, what is its average acceleration during that contact?


    2. Relevant equations



    3. The attempt at a solution

    v1 = -sqrt(-2(g)(4.93m)) = -9.83 m/s
    v2 = sqrt(2(g)(-2.68+4.93)) = 6.64 m/s

    Average acel = v2-v1/delta time = (6.64-(-9.83))/12.3x10^(-3)ms = 1.34 m/s^2?

    The answer isn't correct but I have no idea where I'm going wrong.
     
  2. jcsd
  3. Jan 23, 2008 #2

    G01

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    Your value for v2 is wrong. I think the problem is your value for the height in the v2 equation? Why are you using (-2.68+4.93) as the height in that equation?
     
  4. Jan 23, 2008 #3
    From sqrt(2g(yFinal-yInitial)
     
  5. Jan 23, 2008 #4

    G01

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    Think of the ball traveling back up after the bounce as a completely separate part of the problem from the trip downward before the bounce, with it's own initial and final positions.

    Now, thinking like this, what is the balls initial position after the bounce? What is its final position? Does this change your answer?
     
  6. Jan 23, 2008 #5
    0-(-2.68)?
     
  7. Jan 23, 2008 #6

    G01

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    Yes, that will give you the correct change in position for the ball after the bounce, though your sign choices are confusing me a tiny bit.

    The ball will start at 0 after the bounce, and end at +2.68, thus, yf-yi=2.68.

    This is what you got as well, so it's fine. I am just wondering why you used the signs you did.
     
  8. Jan 23, 2008 #7
    I still end up getting an incorrect answer :( So frustrating!
     
  9. Jan 23, 2008 #8

    G01

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    What do you get for v2 this time around? ( Please show the calculation.)
     
  10. Jan 23, 2008 #9
    sqrt(2*(9.8)(2.68)) = 7.247620299
     
  11. Jan 23, 2008 #10

    G01

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    Ok. This is what I get. Now what do you get for the avg acceleration, calculation included?
     
  12. Jan 23, 2008 #11
    ((7.247620299+9.829954222)/(.00123m/s^2)) = 13884.20693 m/s^2. But its wrong Q_Q
     
  13. Jan 23, 2008 #12

    G01

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    I get 13878 (rounded of course), not 13884. I used those numbers, though. Check your calculations again.
     
  14. Jan 23, 2008 #13
    Rechecked get the exact same answer.
     
  15. Jan 24, 2008 #14

    mda

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    Have you converted the time correctly?
     
  16. Jan 24, 2008 #15

    G01

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    Ahh, yes. I see it now too. Check your time conversion. I think you have a decimal place error. Nice catch mda.
     
  17. Jan 24, 2008 #16
    Always something simple! Raaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa. Thanks for the help! Really appreciate it.
     
  18. Jan 24, 2008 #17

    G01

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    Anytime!
     
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