I just wanna check my work on this problem

[DivGradCurl]

\textrm{Hello, folks. I just wanna check my work on this problem. Thanks.} :cool:

\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
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
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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- Warren