Chapter 10.1, Problem 49E

To determine

To find: Total time travelled by ball.

Answer to Problem 49E

Total time travelled by ball is $T=7.11\text{s}$

Explanation of Solution

Given:

Height from which ball is dropped is 4 m.

Ball rebounds to a height 0.06 h after striking the floor.

Formula used:

A geometric progression with initial term $a$ and common ratio $r$ , Then the sum of this geometric progression with $r<1$ is,

  $S=\frac{a}{1-r}$

Calculation:

Time taken to strike the pavement from maximum height is $\sqrt{\frac{h}{4.9}}\text{s}$

And time taken by ball to reach maximum height is $\sqrt{\frac{h}{4.9}}\text{s}$

Hence, total time taken by the ball is

  $T=\sqrt{\frac{4}{4.9}}+2\left(\sqrt{\frac{4\times 0.06}{4.9}}\right)+2\left(\sqrt{\frac{4\times {0.6}^2}{4.9}}\right)+2{\left(\sqrt{\frac{4\times {0.6}^3}{4.9}}\right)}^3+\mathrm{..}$

Taking,

  $S=2\left[\left(\sqrt{\frac{4\times 0.06}{4.9}}\right)+\left(\sqrt{\frac{4\times {0.6}^2}{4.9}}\right)+{\left(\sqrt{\frac{4\times {0.6}^3}{4.9}}\right)}^3+\mathrm{..}\right]$

Here, it is a geometric progression with initial term $a=2\left(\sqrt{\frac{4\times 0.6}{4.9}}\right)$ and common ratio $r=\sqrt{0.6}$

The sum of this geometric progression with $r<1$ is,

  $\begin{array}{l}S=\frac{a}{1-r}\\ =\frac{2\left(\sqrt{\frac{4\times 0.6}{4.9}}\right)}{\left(1-\sqrt{0.6}\right)}\\ =6.21\end{array}$

Hence, total distance travelled is

  $\begin{array}{l}T=\sqrt{\frac{4}{4.9}}+S\\ =0.9+6.21\\ \Rightarrow T=7.11\text{s}\end{array}$

Conclusion: Total time travelled by ball is $T=7.11\text{s}$