Chapter 11.2, Problem 28SSC

To determine

The number of times the ball will bounce before it bounces back up to a height of about $4\text{}\text{m}$ .

Answer to Problem 28SSC

Approximately 3 times

Explanation of Solution

Given:

Initial height from which the ball is dropped, $h_i=8.0\text{m}$

Formula used:

The gravitational potential energy of an object of mass m and at height $h$ above the surface is given by,

  $PE=mgh$

Calculation:

Since the ball is dropped, thus its initial velocity will be $0\text{m/s}$ . Thus, initial its mechanical energy will be due to potential energy only.

Using above formula, initial mechanical energy of the ball will be,

  $E_i=mg{h}_i$

Now, finally its height is expected to be $h_f=4.0\text{m}$ . Let the number of times the ball bounces back be n , then using energy conservation principle it can be written as,

  $E_i-E_f=n\left(\frac{1}{5}{E}_i\right)$

Using given values in above,

  $\begin{array}{c}mg{h}_i-mg{h}_f=n\left(\frac{1}{5}mg{h}_i\right)\\ h_i-h_f=\frac{n{h}_i}{5}\\ 8-4=\frac{n\left(8\right)}{5}\\ n=2.5\end{array}$

Conclusion:

Thus, in 3^rd round, the ball will bounce up to a height of about $4\text{}\text{m}$ approximately.