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Math Help - Can a formula be written for a repetetive problem?

  1. #1
    Junior Member
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    Can a formula be written for a repetetive problem?

    I drop an object from a height = x foot.
    x = 10

    The object rebounds to .75 of it's previous height.

    Find total distance bounces travelled:
    So for this one I said initial drop plus rebound (down and up) = 10 + 15 = 25 foot.

    This process repeats til it comes to rest. The total distance I came up with doing this is 69.99999 feet after about 65 iterations.

    Instead of 65 lines of cacluations what is a formula that models this?
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by orendacl View Post
    I drop an object from a height = x foot.
    x = 10

    The object rebounds to .75 of it's previous height.

    Find total distance bounces travelled:
    So for this one I said initial drop plus rebound (down and up) = 10 + 15 = 25 foot.

    This process repeats til it comes to rest. The total distance I came up with doing this is 69.99999 feet after about 65 iterations.

    Instead of 65 lines of cacluations what is a formula that models this?
    x + 2x\sum_{k=1}^{\infty} 0.75^k

    Next post has explanation.
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  3. #3
    Super Member Deadstar's Avatar
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    Consider that each bounce has total 'distance' of height multiplied by 2. (since it has to get to the highest point before it drops back down...)

    At each step we are 0.75 the height of the previous step.

    Take initial height to be x.

    Then we have x (distance travelled falling from initial height) + sum of all the bounce distances.

    = x + 2*sum of all the max heights that each bounce reaches.

    = x + 2*(0.75x + 0.75*0.75*x + 0.75*0.75*0.75x + ...)

    = x + 2x*(0.75 + 0.75*0.75 + 0.75*0.75*0.75 + ...)

    = x + 2x\sum_{k=1}^{\infty} 0.75^k.

    And just so you know.

    \sum_{k=1}^{\infty} 0.75*k = \frac{0.75}{1-0.75} = 3

    by the geometric series. (Geometric progression - Wikipedia, the free encyclopedia)

    Hence the whole thing simplifies down to...

    Total distance = 7x


    A final note.

    To generalize this for any initial height and rebound height...

    Initial Height + 2(Initial Height) \bigg{(}\frac{\textrm{how much the object rebounds relative to initial height}}{1 - \textrm{how much the object rebounds relative to initial height}}\bigg{)}
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