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kenreynera
Tyro
Asked: 2021-03-29T19:43:38+00:00 In: ,

geometric sequence

A ball is thrown vertically upward to a height of 45 feet. Each time the ball strike the ground, its rebound is 4/5 of the previous height. How far has the ball travelled when it strikes the ground the sixth time?

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  1. This answer was edited.

    The scenario painted in the question can be modeled as a geometric series with an initial height (term) and common ratio of \frac{4}{5}. We are to find the sixth term.

    The sum of the first 6 terms of a geometric sequence is given by S_n=a\dfrac{1-r^n}{1-r}, where a is the first term and r is the common ratio.

    Thus, S_6=45\times\dfrac{1-\left(\frac{4}{5}\right)^6}{1-\frac{4}{5}}

    =45\times\dfrac{1-0.262144}{0.2}

    =45\times\dfrac{0.737856}{0.2}

    =45\times3.68928

    =166.0176

    Now, we take notice that for each height, there is a rise and there is a fall. Thus, for each height, there the ball travels twice the height.

    Therefore, the total distance traveled by the ball by the time it hits the ground the sixth time is 2\times166.0176=332.0352 feet.

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