Chapter 9.3, Problem 90E

To determine

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

Answer to Problem 90E

The total vertical distance the ball travels before rest is $S=42\text{ feet}$ .

Explanation of Solution

Given information:The height of each bounces is three − fourths the height of the previous bounce and graph

  

Formula needed:sum of infinite geometric sequence $\left(a{r}^n\right)$ is $S=\frac{a}{1-r},r<1\text{ and }a\text{ is first term}$

Calculation:Initially, the vertical distance covered by ball is 6 feet i.e., $a_1=6$

After first bounce, the vertical distance covered by ball is two times of previous height i.e., $a_2=\frac{3}{4}\cdot a_1+\frac{3}{4}\cdot a_1=2\cdot \left(\frac{3}{4}{a}_1\right)$

After second bounce, the vertical distance covered by ball is two times of previous height i.e., $a_3=\frac{3}{4}\left(\frac{3}{4}\cdot a_1\right)+\frac{3}{4}\left(\frac{3}{4}\cdot a_1\right)=2{\left(\frac{3}{4}\right)}^2\cdot a_1$

After third bounce, the vertical distance covered by ball is two times of previous height i.e., $a_4=\frac{3}{4}\left(\frac{3}{4}\left(\frac{3}{4}\cdot a_1\right)\right)+\frac{3}{4}\left(\frac{3}{4}\left(\frac{3}{4}\cdot a_1\right)\right)=2{\left(\frac{3}{4}\right)}^3\cdot a_1$

And so on.

The n^th bounce is given by

  $a_n=2\cdot {\left(\frac{3}{4}\right)}^{n-1}{a}_1$

The total vertical distance is $S=a_1+a_2+a_3+\cdot \cdot \cdot +a_n+\cdot \cdot$

  $\begin{array}{l}S=a_1+2\left(\frac{3}{4}{a}_1\right)+2{\left(\frac{3}{4}\right)}^2{a}_1+\cdot \cdot \cdot +2{\left(\frac{3}{4}\right)}^{n-1}{a}_1+\cdot \cdot \\ S=a_1+\frac{3}{2}{a}_1\left[1+\frac{3}{4}+{\left(\frac{3}{4}\right)}^2+\cdot \cdot \cdot +{\left(\frac{3}{4}\right)}^{n-2}+\cdot \cdot \right]\\ S=a_1+\frac{3}{2}{a}_1\left(\frac{1}{1-0.75}\right)\\ S=a_1+\frac{3}{2}{a}_1\left(\frac{1}{0.25}\right)\\ S=a_1+\frac{3}{2}{a}_1\left(4\right)\\ S=a_1+6a_1\\ S=7a_1\end{array}$

Here, $a_1=6$

  $\begin{array}{l}S=7\cdot 6\\ S=42\end{array}$

Hence, the total vertical distance the ball travels before rest is $S=42\text{ feet}$ .