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Geometric progression ball drop problem

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball is dropped from a height of 24m and rebounds to a height of 16m. If each time it rebound two-thirds of the previous height, find the total distance travelled by the ball.


    2. Relevant equations



    3. The attempt at a solution
    I thought this might be a problem approaching infinity even though the ball cannot bounce infinitely. So i used the equation a/1-r where r was two-thirds and a was 24. That didn't work; however, when I added 24 and 16 then divided it by 1-r i got the correct answer, 120m. Was this a correct procedure? if not could you explain which procedure should be used in this case.
     
  2. jcsd
  3. Apr 18, 2012 #2
    Hey, 24m was the height the ball dropped at 16m was the height it rised. From then on the height dropped=previous height rised=2/3 previous height dropped and height rised=2/3 previous height rised.

    So i got two sum of infinities,
    Distance travelled = 24/1-(2/3) + 16/1-(2/3) = 120
     
  4. Apr 18, 2012 #3

    Curious3141

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    Homework Helper

    Your second procedure is right, but it looks like you stumbled on it, rather than figured it out.

    Each bounce of the ball takes it to 2/3 the previous height reached. But the ball also needs to come down from that height to hit the floor before it bounces again.

    So the balls falls from 24m, rebounds to 16m, falls 16m, rebounds to (16*2/3)m, falls (16*2/3)m, ...etc.

    Hence you're really adding up two infinite geometric series, both with common ratio 2/3. The first denotes the "fall" steps and goes 24, 16, (16*2/3),... and the second denotes the "rise" steps and goes 16,(16*2/3),...

    The total distance travelled is the sum of those two series, and it's essentially [itex]\frac{a_1}{1-r} + \frac{a_2}{1-r} = \frac{a_1 + a_2}{1-r}[/itex], which is why adding 24 and 16, then dividing by (1 - 2/3) works here.
     
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