Bouncing ball: total distance, time, and time to rest

[DivGradCurl]

$$\textrm{Hello, folks. I just want to check my work on this problem. Thanks.}$$

$$\textrm{A certain ball has the property that each time it falls from a height}$$ $$h$$ $$\textrm{onto a hard, level surface, it rebounds to a height}$$ $$rh$$ $$\textrm{, where}$$ $$0<r<1$$. $$\textrm{Suppose that the ball is dropped from an initial height of}$$ $$H$$ $$\textrm{meters.}$$

$$\textrm{(a) Assuming that the ball continues to bounce indefinitely, find the total distance that<br /> it travels.}$$

$$H + 2rH + 2r^{2}H + 2r^{3}H + \cdots = H + 2H \sum _{n=1} ^{\infty} \left( r \right) r^{n-1} = H + 2H \left( \frac{r}{1-r} \right) = H \left( \frac{1+r}{1-r} \right)$$

$$\textrm{(b) Calculate the total time that the ball travels.}$$

$$t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + \sqrt{\frac{2rH}{g}} + \sqrt{\frac{2r^2H}{g}} + \sqrt{\frac{2r^3 H}{g}} + \cdots$$

$$t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + \sqrt{\frac{2rH}{g}} \left( 1 + \sqrt{r} + \sqrt{r^2} + \sqrt{r^3} + \cdots \right)$$

$$t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + \sqrt{\frac{2rH}{g}} \left( \frac{1}{1-\sqrt{r}} \right)$$

$$\textrm{(c) Suppose that that each time the ball strikes the surface with velocity}$$ $$v$$ $$\textrm{it rebounds with velocity}$$ $$-kv$$$$\textrm{, where}$$ $$0<k<1$$. $$\textrm{How long will it take for the ball to come <br /> to rest?}$$

$$v_{\textrm{REST}} = v + kv + k^2 v + k^3 v + \cdots$$

$$v_{\textrm{REST}} = v + \sum _{n=1} ^{\infty} \left( k v \right) k^{n-1}$$

$$v_{\textrm{REST}} = v + \left( \frac{kv}{1-k} \right)$$

$$\textrm{If } K=U, \textrm{we find}$$

$$\frac{1}{2}mv_{\textrm{REST}} ^2= mgH$$

$$\frac{1}{2}m\left[ v^2 + 2v^2 \left( \frac{k}{1-k} \right) + v^2 \left( \frac{k}{1-k} \right)^2 \right] = mgH$$

$$H = \frac{1}{2g}\left[ v^2 + 2v^2 \left( \frac{k}{1-k} \right) + v^2 \left( \frac{k}{1-k} \right)^2 \right]$$

$$\textrm{which gives}$$

$$t_{\textrm{REST}} = - \frac{2\left| \frac{v}{g\left( k-1 \right)} \right|}{\sqrt{r}-1} - \left| \frac{v}{g\left( k-1 \right)} \right|$$

 

[Galileo]

At b)
The time it takes to fall from a height h is $\sqrt{\frac{2h}{g}}$, so the time it takes to rebound to a height h and fall down to the ground again is twice as long.

$$t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} + 2\sqrt{\frac{2r^2H}{g}} + 2\sqrt{\frac{2r^3 H}{g}} + \cdots$$

 

[DivGradCurl]

$$\textrm{Yes, indeed. I should have written}$$

$$t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} + 2\sqrt{\frac{2r^2H}{g}} + 2\sqrt{\frac{2r^3 H}{g}} + \cdots$$

$$t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} \left( 1 + \sqrt{r} + \sqrt{r^2} + \sqrt{r^3} + \cdots \right)$$

$$t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} \left( \frac{1}{1-\sqrt{r}} \right)$$

$$\textrm{Thanks.}$$

 

[DivGradCurl]

Guys, I'm not so sure about what I found for part (c). Did I get it right?

Thank you.​

 

[chroot]

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