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'