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Q. 80

Expert-verifiedFound in: Page 616

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Whenever a certain ball is dropped, it always rebounds to a height *p% (0 < p < 100)* of its original position. What is the total distance the ball travels before coming to rest when it is dropped from a height of *h* meters?

The total distance the ball travels before coming to rest when it is dropped from a height of *h *meters is $h+\frac{ph}{50}\left(\frac{100}{100-p}\right).$

The ball after dropping rebounds to a height *p**% (0 < p < 100)* of its original position.

To find the total distance traveled by the ball, let's first find the distance traveled by the ball before coming to rest, each successive bounce has two times *p**% *of the height as it involves up and down of the ball,

$d=h+2\times h\times \left(\frac{p}{100}\right)+2\times h\times {\left(\frac{p}{100}\right)}^{2}+...\phantom{\rule{0ex}{0ex}}d=h+2\times h\times \frac{p}{100}\left(1+\left(\frac{p}{100}\right)+{\left(\frac{p}{100}\right)}^{2}+...\right)\phantom{\rule{0ex}{0ex}}\mathrm{Use}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{geometric}\mathrm{series}\left(\frac{\mathrm{a}}{1-\mathrm{r}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{d}=h+2\times h\times \frac{p}{100}\left(\frac{1}{1-\frac{p}{100}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{d}=h+\frac{ph}{50}\left(\frac{100}{100-p}\right)$

Thus, the total distance traveled by the ball before coming to the rest is $h+\frac{ph}{50}\left(\frac{100}{100-p}\right).$

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