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A ball is dropped from a height of 9 metres and rebounds to a height of 6 metres and then on each bounce rises to $\frac{2}{3}$ of the previous height. Find the total distance through which the ball moves before coming to rest.

I understand this question is about geometric series.

Thus $S_{\infty} = \frac{9}{1-2/3} = 27$

But the solution gives total distance = 2*27-9 = 45m

Why is not the answer 27m?

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The ball has to bounce both up and down on every bounce, except the initial drop. So, you get $2$ times $27$ to double all the drops. But, you have to subtract $9$ because the initial drop does not include an "up" component.

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  • $\begingroup$ wow, I feel silly now. Thanks joe, for pointing out what was an easy answer. hehe. $\endgroup$
    – yiyi
    May 20, 2012 at 11:43

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