Chapter 8.2, Problem 52E

a.

To determine

To find: the total distance the ball travels assuming that the ball continues to bounce indefinitely.

Answer to Problem 52E

  $D=\frac{h(1+r)}{1-r}$ .

Explanation of Solution

Given information: A certain ball has the property that each time it fallsfrom a height h

onto a hard ,level surface , it rebounds to a height rh , where 0<r < 1. Suppose that the ball is dropped from an initial height of H meters.

Calculation:

The ball falls an initial height of H meters.

Then climbs a height of rH meters.

Then falls a height rH meters.

Then climbs a height of $r^{2}H$ meters.

Then falls a height $r^{2}H$ meters.

Then climbs a height of $r^{3}H$ meters.

Then falls a height $r^{3}H$ meters.

And so on…..

Total distance travelled by the ball is:

  $\begin{array}{l}D=H+rH+rH+r^{2}H+r^{2}H+r^{3}H+r^{3}H+\mathrm{.....}\\ D=H+2rH+2r^{2}H+2r^{3}H+\mathrm{.....}\\ D=[{\displaystyle \sum _{n=0}^{\infty }2H{r}^n}]-H\\ D=2H[{\displaystyle \sum _{n=0}^{\infty }{r}^n}]-H\\ D=2H.\frac{1}{1-r}-H.\\ D=\frac{2H-H+Hr}{1-r}\\ D=\frac{h(1+r)}{1-r}.\end{array}$

b.

To determine

To calculate: the total time that the ball travels.

Answer to Problem 52E

  $T=\sqrt{\frac{2H}{g}}.\frac{1+\sqrt{r}}{1-\sqrt{r}}.$

Explanation of Solution

Given information: A certain ball has the property that each time it fallsfrom a height h

onto a hard ,level surface , it rebounds to a height rh , where 0<r < 1. Suppose that the ball is dropped from an initial height of H meters.

Calculation:

Calculate the length of each fall, then add them. For each n , need to caalaculate the time $t_n$ it takes to fall $H{r}^n$ meters.

  $\begin{array}{l}\frac{1}{2}g{t}^2=H{r}^n\\ t_n{}^2=\frac{2H}{g}{r}^n\\ $ t_n=\sqrt{\frac{2H}{g}}{(\sqrt{r})}^n$\end{array}$

Noting that it takes the same amount of time to climb $H{r}^n$ meters as it does to fall that far,

  $\begin{array}{l}$ T=t_0+2t_1+2t_2{}^2+\mathrm{.....}={\displaystyle \sum _{n=0}^{\infty }2\sqrt{\frac{2H}{g}}}{(\sqrt{r})}^n-t_0$$ ={\displaystyle \sum _{n=0}^{\infty }2\sqrt{\frac{2H}{g}}}{(\sqrt{r})}^n-\sqrt{\frac{2H}{g}}=\sqrt{\frac{2H}{g}}(\frac{2}{1-\sqrt{r}}-1)=\sqrt{\frac{2H}{g}}.\frac{1+\sqrt{r}}{1-\sqrt{r}}$T=\sqrt{\frac{2H}{g}}.\frac{1+\sqrt{r}}{1-\sqrt{r}}.\end{array}$

c.

To determine

To find: how long will it take for the ball to come to rest. Suppose that each time the ball strikes the surface with velocity v it rebounds with velocity − kv , where 0

Answer to Problem 52E

  $T=\frac{v}{g}.\frac{1+k}{1-k}.$

Explanation of Solution

Given information: A certain ball has the property that each time it fallsfrom a height h

onto a hard ,level surface, it rebounds to a height rh , where 0<r < 1. Suppose that the ball is dropped from an initial height of H meters.

Calculation:

  $g{t}_n=k^{n}v\Rightarrow t_n=k^n\frac{v}{g}$ .

Now add:

  $\begin{array}{l}T=t_0+2t_1+2t_2+\mathrm{....}={\displaystyle \sum _{n=0}^{\infty }\frac{2v}{g}}{k}^n-\frac{v}{g}\\ T=\frac{v}{g}({\displaystyle \sum _{n=0}^{\infty }2}{k}^n-1)=\frac{v}{g}(\frac{2}{1-k}-1)=\frac{v}{g}.\frac{1+k}{1-k}.\\ T=\frac{v}{g}.\frac{1+k}{1-k}.\end{array}$