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 the ball travels before coming to rest is 42 feet.

Explanation of Solution

Given information: A ball is dropped from a height of 6 feet and begins bouncing. The

height of each bounce is three-fourths the height of the previous bounce

Calculation:

The ball falls 6 feet, then rebounds to a height $\frac{3}{4}(6)=4.5$ feet, then falls 4.5 feet, then rebounds $\frac{3}{4}(4.5)=3.375$ feet, and so on .

The terms of this geometric series can be written as

6 + (4.5) (2) + (3.375) (2) +.......or

  $\begin{array}{c}6+9+\frac{3}{4}(9)+{\left(\frac{3}{4}\right)}^2(9)+\mathrm{.......}={\displaystyle \sum _{}^{}9{\left(\frac{3}{4}\right)}^{n-1}}\\ \text{So,}\text{the}\text{sum}\text{is}:\\ 6+\frac{9}{1-\frac{3}{4}}=42\text{feet}\text{.}\end{array}$

Therefore, the total vertical distance the ball travels before coming to rest is 42 feet.