When Does a Bouncing Ball Come to Rest?

[DeldotB]

Homework Statement

Hello!
A ball is dropped and falls to the floor (no horizontal velocity). It hits the floor and bounces with inelastic collisions. The velocity after each bounce is $\mu$ times the velocity of the previous bounce (here $\mu$ is the constant of restitution). The velocity of the first bounce is just $v_0$. Find the time it takes for the ball to stop bouncing.

Homework Equations

Newtons Laws

The Attempt at a Solution

Well:
I know this will turn into a convergent geometric series. I am just trying to find what that series will look like.

using the formula $h=x_0+v_0t+1/2at^2$ its easy to see that the time it takes for the ball to reach the ground is:

$h=1/2gt^2$ so $t=\sqrt{2h/g}$.
Using energy I also have: $mgh=1/2mv_0^2$ so $gh=1/2v_0^2$

Time for the next bounce: well, the ball now has an upward velocity of $\mu v_0$ and the height of the first bounce is $h'=\mu v_0t-1/2gt^2$.

I realize this is a simple problem but for some reason I'm not seeing it. If I solve this equation for time, (using quadratic formula) the resulting series for the times $t=t_1+t_2+...$ isn't geometric and actually quite complicated. Is my approach right?

 

[BvU]

Hi,
You don't want the height of the next bounce, but the time for the ball to reach the ground again. Easier to solve, too!

 

[DeldotB]

I solved the quadratic for the time. Is this not the right approach?

 

[BvU]

Should lead to the same answer - with a lot more work.
What did you use for $h'$ ?

 

[DeldotB]

Should lead to the same answer - with a lot more work.

Ahh, I see what you mean. So after the first bounce, I have:

Time for ball to reach ground again: $0= \mu v_0 -1/2gt^2$ solvig for t yields: $2 \mu v_0/g$ so

$t_1 =2 \mu v_0/g$

Time for ball the reach the ground the third time:

$t_2= 2 (\mu)^2 v_0/g$
and so on. Is this the right direction?

 

[BvU]

And there you have your geometric sequence !
Make sure you have the right summation: the first t is only 'half a bounce'