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Bouncing ball series

  1. Jul 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball is dropped from a height of 16 feet. Each time it drops h feet, it rebounds 0.81h feet. Find the total distance traveled by the ball.

    2. Relevant equations

    Geometric series sum of ar^n

    3. The attempt at a solution

    D1= 16 feet
    D2 = 16(0.81) + 16(0.81) = 32(0.81)
    Then D3 = 32(0.81)^2

    So then you have D = 16 + 32(sum from n = 1 to ∞) (0.81)^n+1

    = 16 + 32(0.81)(sum from n = 0 to ∞) (0.81)^n
    = 152.42 ft (right answer)
    When I first did my own I got D = 16 + 32(sum from n = 0 to ∞)(0.81)^n

    My question is why is this not correct?


    Also the book did it a different way and had
    -16 + (sum from n = 0 to ∞)32(0.81)^n

    why would you subtract 16? This is what I had when it was wrong except they subtract 16. Why?
     
  2. jcsd
  3. Jul 4, 2013 #2

    Nugso

    User Avatar
    Gold Member

    Hello. Sorry but I didn't quite understand. ( That's probably because of my English)

    The way your book answers it is correct. Because by summing 32(0,81)^n where n is from 0 to ∞, it assumes that the ball travels D1 twice. ( Up and down)

    Here is how it looks like without subtracting 16;

    http://www.sketchtoy.com/42921712

    Here is the exact answer;

    http://www.sketchtoy.com/42921326

    We have D= 16 + 32(sum from n=1 to ∞)(0,81)n
     
    Last edited: Jul 4, 2013
  4. Jul 4, 2013 #3
    Because the ball only has one trip of length 16. If the ball started from the ground and bounced up 16 feet, then the total distance would be [itex] 2\sum_{n=0}^{\infty} 16(.81)^n= \sum_{n=0}^{\infty} 32(.81)^n[/itex]. The answer is 16 less than this since the first trip of length 16 is not made.
     
  5. Jul 4, 2013 #4
    Oh I see thanks
     
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