Chapter 8.3, Problem 111E

To determine

To find: the total vertical distance the ball travels before coming to rest.

Answer to Problem 111E

The total vertical distance travelled by the ball before it comes to rest is 24 feet.

Explanation of Solution

Given information:

A ball is dropped from a height of 6 feet and begins bouncing, as shown in the figure. The height of given bounce is three fourths the height of the previous bounce.

Given figure

  

Calculation:

It is stated that a ball is dropped from a height of 6 feet and begins bouncing. The height of given bounce is three fourths the height of the previous bounce.

The sequence of heights are given by $6,6{\left(\frac{3}{4}\right)}^1,6{\left(\frac{3}{4}\right)}^2,6{\left(\frac{3}{4}\right)}^3,\mathrm{...},6{\left(\frac{3}{4}\right)}^{n-1},\mathrm{...}$ which is nothing but a geometric sequence with first term being $a_1=6$ and the common ratio $r=\frac{3}{4}$ respectively.

To determine the total vertical distance travelled by the ball before it comes to rest we have to apply the formula to find the sum of an infinite geometric series because for the following reason.

  $\begin{array}{c}{a}_n=a_1{r}^{n-1}\\ a_n=6{\left(\frac{3}{4}\right)}^{n-1}\\ 0=6{\left(\frac{3}{4}\right)}^{n-1}\\ {\left(\frac{3}{4}\right)}^{n-1}=0\\ n\to \infty \end{array}$

Now determine the total vertical distance travelled by the ball before it comes to rest as shown below:

  $\begin{array}{c}S={\displaystyle \sum _{n=1}^{\infty }{a}_1{r}^n}\\ =\left(\frac{a_1}{1-r}\right)\\ =\frac{6}{1-\left(\frac{3}{4}\right)}\\ =\frac{6}{\left(\frac{1}{4}\right)}\\ =24\end{array}$

Thus, the total vertical distance travelled by the ball before it comes to rest is 24 feet.