Question

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?

Short answer

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

Step-by-step solution

Step 1. Given Information. 

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

Step 2. Find the distance the ball travels before coming to rest when it is dropped from a height of h meters. 

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,

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

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