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Is there an easy way to calculate this problem?

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A Bouncy ball is dropped from a window, 28.7 meters above the ground. It bounces inelastically off the ground and bounces up again, but not as high as it stated out. If it looses 9 percent of its kinetic energy every time it hits the ground, how many times can it bounce, and still have a maximum height greater than half the original window height.

    2. Relevant equations
    1/2 mv^2 = mgh

    3. The attempt at a solution
    Since I know it loses 9% of its kinetic energy, the potential will be lowered by 9% each time the ball bounces also.

    So I did 28.7 - (0.09 x 28.7) = 26.1

    then 26.1 - (0.09 x 2.61) = 23.76

    then 23.76 - (0.09 x 23.76) = 21.627

    and so on, I keep repeating this until I get an answer below 28.7 / 2 = 14.35 meters.

    And after all that, I get the answer to be 7 bounces.


    Is there a faster way to do this problem?
     
  2. jcsd
  3. Mar 20, 2015 #2

    ehild

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    The question is, how many times can it bounce, and still have a maximum height greater than half the original window height.
    The seventh bounce produces less height than half the initial. Count the bounces till the height is still greater than half the initial height, but the next bounce produces less than the half.
    What is the relation between the k-th height and the k+1 height? What sequence do the heights make?
     
  4. Mar 20, 2015 #3
    what is k-th and k+1?
     
  5. Mar 20, 2015 #4

    ehild

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    k represents the k-th bounce. It is just a positive integer. For example, the height after the third bounce is 0.91 times the height after the second bounce.
     
    Last edited: Mar 20, 2015
  6. Mar 20, 2015 #5
    the previous bounce is always 1.01 times higher.

    28.7/26.11 = 1.01

    26.11/23.76= 1.01

    23.76/21.62 = 1.01
    etc etc

    so I can just keep dividing by 1.01 to get a new height.
     
  7. Mar 20, 2015 #6

    ehild

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    Why do you divide by 1.01? If the energy loss is 9 percent it means that the energy after the next bounce is 0.91 times the previous one. Dividing by 1.01 produces 0.99 times the previous one, which is 1 percent loss.

    The next bounce is 0.91 times the previous one. Considering the heights as a sequence h0, h1, h2... hk what is the relation between one term and the next one? What kind of sequence is it?
     
  8. Mar 20, 2015 #7
    oh, im sorry, I meant 1.1, not 1.01.
     
  9. Mar 20, 2015 #8

    ehild

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    Instead of dividing, use multiplication. Loosing 9 % means that 91% remains. Dividing by 1.1 means that 90.90909...% remains.They are not the same.
    So how do you get the height after the first bounce from the initial height? How do you get the height after the second bounce?
     
  10. Mar 20, 2015 #9
    first height times 0.91 gives new height , then we just keep repeating?
     
  11. Mar 20, 2015 #10

    ehild

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    Keeps repeating. How many times is the second height of the initial height? How many times is the third height of the initial height? How many times is the k-th height of the initial height?
     
  12. Mar 20, 2015 #11
    ah
    the first bounce is 0.91 times the initial height, and the 2nd bounce is 0.82 times the initial , and the 3rd bounce is .73 the initial.

    I'm assuming k-th is the height we want it at, which is 14.35 which is .5 times the initial.
     
  13. Mar 20, 2015 #12

    ehild

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    How is it symbolically? If the initial height is H the first is
    h1= H*0.91,
    the second is
    h2=H*0.91*0.91=H*0.912,
    the third is
    h3=H*0.913,
    the k-th is
    hk=H*0.91k.

    So you have the relation

    hk ≥ 0.5 .

    What is the maximum value of k so as the inequality is valid? How do you express k with logarithm?
     
  14. Mar 20, 2015 #13
    log(14.35)/log(28.7) = .7935

    is that it?
     
  15. Mar 20, 2015 #14

    ehild

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    No. Take the logarithm of both sides what do you get?
     
  16. Mar 20, 2015 #15
    you mean log h^k ≥ log 0.5

    ???
     
  17. Mar 20, 2015 #16

    ehild

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    Yes. Do you know how to take the logarithm of something on the k-th power?
     
  18. Mar 20, 2015 #17
    Sadly, no. I will just look it up.
     
  19. Mar 20, 2015 #18

    ehild

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    Haven't you studied logarithm yet?
     
  20. Mar 20, 2015 #19
    yes, but it was a while ago and somethings have been forgotten
     
  21. Mar 20, 2015 #20

    ehild

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    log(a b) = log (a) +log(b)
    log(a/b) = log (a)-log (b)

    log (ak) = k log(a) ... :smile:
     
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