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Math Help - [SOLVED] Geometric Series

  1. #1
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    [SOLVED] Geometric Series

    A ball is dropped from a height of 10m. Each time it strikes the ground it bounces vertically to a height that is 3/4 of the preceding height. Find the total distance the ball will travel if is assumed to bounce infinitely often

    I get 40m as my answer but the back of the book says 70m
    this is what im doing.
    \sum_{k=0}^{\infty}10*(\frac{3}{4})^k
    with r=\frac{3}{4};a=10

    \frac{10}{1-\frac{3}{4}}=                \frac{10}{\frac{1}{4}}
    =40m
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  2. #2
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    Quote Originally Posted by vinson24 View Post
    A ball is dropped from a height of 10m. Each time it strikes the ground it bounces vertically to a height that is 3/4 of the preceding height. Find the total distance the ball will travel if is assumed to bounce infinitely often

    I get 40m as my answer but the back of the book says 70m
    this is what im doing.
    \sum_{k=0}^{\infty}10*(\frac{3}{4})^k
    with r=\frac{3}{4};a=10

    \frac{10}{1-\frac{3}{4}}=                \frac{10}{\frac{1}{4}}
    =40m
    Were you giving that summation or did you have to come up with it yourself?
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  3. #3
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    You are not taking into account the distance traveled when the ball is rising. In other words, when it strikes the ground, it bounces up and then covers that distance again going back down. Since it rises and falls the same amount, you can double your answer to get 80m. But when it is initially dropped, there was not a "rise" of 10m, so we have overcounted by 10m. This gives 70m.

    (In other words, the correct sum is  10 + 20\sum_1^\infty (\frac34)^k.)
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  4. #4
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    i came up with the summation but thanks i didnt even think of that
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  5. #5
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    Quote Originally Posted by maddas View Post
    You are not taking into account the distance traveled when the ball is rising. In other words, when it strikes the ground, it bounces up and then covers that distance again going back down. Since it rises and falls the same amount, you can double your answer to get 80m. But when it is initially dropped, there was not a "rise" of 10m, so we have overcounted by 10m. This gives 70m.

    (In other words, the correct sum is  10 + 20\sum_1^\infty (\frac34)^k.)
    That sum adds up to 90.

    Shouldn't be  20\sum_{n=0}^\infty (\frac{3}{4})^n-10
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  6. #6
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    We've written the same sum! (Look at the index on my summation ;])
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  7. #7
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    I see I see, my bad.
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