A ball is dropped from a height of 12m. Each time it strikes the ground, it bounces back to a height three-fourths the distance from which it fell. Find the total distance traveled by the ball. See Figure 18.

Solution Let $h_{n}$ denote the height of the ball on the $n$th bounce. Then $$\begin{array}{cl} h_{0} & =12 \\ h_{1} & =\dfrac{3}{4}(12) \\ h_{2} & =\dfrac{3}{4}\left[ \dfrac{3}{4}(12)\right] =\left( \dfrac{3}{4}\right) ^{2}(12) \\ \vdots & \\ h_{n} & =\left( \dfrac{3}{4}\right) ^{n}(12) \end{array}$$

After the first bounce, the ball travels up a distance $h_{1}=\dfrac{3}{4}(12)$ and then the same distance back down. Between the first and the second bounce, the total distance traveled is therefore $h_{1}+h_{1}=2h_{1}$. The total distance $H$ traveled by the ball is $$\begin{eqnarray*} H=h_{0}+2h_{1}+2h_{2}+2h_{3}+\cdots &=& h_{0}+\sum\limits_{k\,=\,1}^{\infty }\left( 2h_{k}\right) =12+\sum\limits_{k\,=\,1}^{\infty }2 \left[ 12 \left( { \frac{3}{4}}\right) ^{k}\right]\\[8pt] &=& 12+\sum\limits_{k\,=\,1}^{\infty }24\left[ \frac{3}{4}\left( {\frac{3}{4}}\right) ^{k-1}\right] \\ &=& 12+\sum\limits_{k\,=\,1}^{\infty }18\left( {\frac{3}{4}}\right) ^{k-1} \end{eqnarray*}$$

The sum is a geometric series with $a=18$ and $r=\dfrac{3}{4}$. The series converges and $$H=12+\sum\limits_{k\,=\,1}^{\infty }18\left( \frac{3}{4}\right) ^{k-1}=12 + \frac{18}{1-\dfrac{3}{4}}=84$$

The ball travels a total distance of $84$ m.