Chapter 12.3, Problem 82E

To determine

To estimate the time interval from the instant the ball first touches the ground until it stops bouncing.

Answer to Problem 82E

  $\left[\frac{1}{2},2+\sqrt{2}\right)$

Explanation of Solution

Given:

The ball is dropped form a height of 8 ft and takes 1 s to complete the first bounce. Each bounce requires $\frac{1}{\sqrt{2}}$ as long as the preceding complete bounce.

Calculation:

The ball touches the ground for the first time after $\frac{\left(\frac{1}{\sqrt{2}}\cdot \sqrt{2}\right)}{2}=\frac{1}{2}\text{s}$ .

For the given information, the infinite geometric series of the time required to complete the bounce will be $1,\frac{1}{\sqrt{2}},{\left(\frac{1}{\sqrt{2}}\right)}^2,{\left(\frac{1}{\sqrt{2}}\right)}^3,\cdot \cdot \cdot$

So, the total time required for the ball to come to rest is given by the sum of the infinite series.

  $\begin{array}{c}S=\frac{1}{1-\frac{1}{\sqrt{2}}}\\ =\frac{\sqrt{2}}{\sqrt{2}-1}\\ =\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\\ =\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{{\left(\sqrt{2}\right)}^2-1^2}\\ =\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{2-1}\\ =\sqrt{2}\left(\sqrt{2}+1\right)\\ =2+\sqrt{2}\end{array}$

So, the time interval for the ball, from the instant it first touches the ground until it stops bouncing is

  $\left[\frac{1}{2},2+\sqrt{2}\right)$ .