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Dropped tennis ball

  1. Oct 14, 2003 #1
    Hello! I was hoping I could receive some help with a certain problem.

    A 0.7kg tennis ball is dropped from rest at a height of 5m onto a hard floor.

    --From that, I was able to determine that the final velocity is 9.9m/s.

    The part I'm having trouble with is this: A flash photograph shows that the ball is compressed a maximum of 0.6cm when it strikes the floor. Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?

    I really don't know what to do with this information...
    Any help would be greatly appreciated!
  2. jcsd
  3. Oct 14, 2003 #2

    Tom Mattson

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    It means that the ball was brought to a stop in a distance of 0.6cm. Use your final velocity of the first part as the initial velocity of the second part (the final velocity of the second part is obviously zero).

    You know:

    *Initial Velocity
    *Final Velocity
    *Change In Displacement

    You need:


    There is precicely one equation that relates all of those. Can you find it?
  4. Oct 14, 2003 #3
    Thank you! I was finally able to solve it...
  5. Oct 14, 2003 #4
    ...And yet I run into another problem.

    After I find the force it asks me, "what time does the force act in bringing the ball to rest?"

    I realize I'm supposed to use one of the kinematic equations, so I've been plugging in numbers for a and x, etc. The thing is, all of my solutions are wrong... I'm guessing that I've been plugging in values that aren't supposed to be plugged in....?

    Help please? Many thanks in advance!
  6. Oct 15, 2003 #5


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    You know the force now so you know the acceleration.

    You know that the speed at time t is given by v(t)= v0+ at
    (a is positive (up) here, v0= -9.9 m/s) and you are looking for the time until v(t)= 0: that is t= -v0/a.)
  7. Oct 15, 2003 #6
    I keep getting the wrong answer...Grr. :frown:

    Earlier, I determined that a=8167.5m/s^2 and f=5717.25N.
    The online program I'm using told me that my answer was correct (for force), so I proceeded to solve for time.

    No matter how many times I plug the numbers into the equation, I keep getting 0.0012s, which is wrong, according to this program.

    Did I make a miscalculation somewhere?
  8. Oct 16, 2003 #7
    My bad.

    The program refused to take 0.0012s, but it took 0.00121s.
    I guess I just needed to be a little more accurate...(--;)
    I'm such an idiot.

    Anyway, thank you for all the help!
    Now I have an exam to look forward to on Friday :)
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