Question

A certain ball has the property that each time it falls from aheight h onto a hard level surface, it rebounds to aheight rh, where 0 < r < 1. Suppose thatthe ball is dropped from an initial height H meters.
a)Assuming that the ball continues to bounceindefinitely, find the total distance that it travels. (Use the fact thatthe ball falls (1/2)gt2 meters in tseconds)
b)Calculate the total time that the balltravels.
c)Suppose that each time the ball strikes thesurface with velocity v it rebounds with velocity -kv, where0 < k < 1. How long will it take for the ball tocome to rest?

Answer 1

Answer:

a) $H(\frac{1+r}{1-r})$

b) $t=\sqrt{\frac{2H}{g}\left ( \frac{1+r}{1-r} \right )}$

c) The ball never gets velocity 0.

Step-by-step explanation:

a)

Assuming the initial height is H meters we have the total distance traveled is  

H + distance traveled 1st bounce +  distance traveled 2nd  bounce+ distance traveled 3rd  bounce +...(and so on)

The distances traveled in each bounce are

$\large \left[\begin{array}{ccc}Bounce&distance\; traveled\\1&2rH\\2&2r^2H\\3&2r^3H\\....&....\\n&2r^nH\\...&... \end{array}\right]$

So, the total distance traveled is

$\large H+2rH+2r^2H+2r^3H+...=H(1+2(r+r^2+r^3+...))$

but  

$\large r+r^2+r^3+...=\sum_{k=1}^{\infty}r^k$

is a convergent geometric series (without the first term=1) which converges to

$\large \frac{1}{1-r}-1$

and we have that the total distance traveled is

$H(1+2(\frac{1}{1-r}-1))=\boxed{H(\frac{1+r}{1-r})}$

b)

Since the total time t traveled by the ball is given by the second equation of motion

$\large H(\frac{1+r}{1-r})=\frac{gt^2}{2}$

we then have

$\large H(\frac{1+r}{1-r})=\frac{gt^2}{2}\rightarrow t^2=\frac{2H}{g}\left ( \frac{1+r}{1-r} \right )$

and finally

$\large t=\sqrt{\frac{2H}{g}\left ( \frac{1+r}{1-r} \right )}$

c)

At bounce number n the velocity would be $\large  -(k)^nv$ and we know  

$\large \lim_{n\to \infty}k^n=0$

for 0 < k < 1.

So, theoretically speaking, the ball will never have velocity zero, but will become smaller and smaller until it becomes zero in the infinite.