# Maximum force and Impulse

1. Feb 28, 2009

### julz3216

1. The problem statement, all variables and given/known data

A 41 rubber ball is dropped from a height of 1.8 and rebounds to two-thirds of its initial height.
http://session.masteringphysics.com/problemAsset/1070447/4/09.P29.jpg
The figure (Part C figure) shows the impulse received from the floor. What maximum force does the floor exert on the ball?
t=5ms

2. Relevant equations

change in p = F*change in time
F=ma

3. The attempt at a solution

I tried and got Fmax = .176 by plugging into the 1st equation for impulse

2. Mar 1, 2009

### LowlyPion

If you solved the other problem you posted you should know how to solve this so long as you calculated your Impulse correctly from the info given.

3. Mar 1, 2009

### julz3216

So I tried to do F*change in t = change in m*v but it didn't work. i calculated velocity by doing change in d/ change in t. i used 2.5 for t because that is where fmax occurs and i also used d as 1.8 because that is the distance the ball travels in 2.5 seconds

so i got v=.72 and then mv=.029
so 2.5F = .029
F = .0116

and that was wrong. so i dont really know what to do?

4. Mar 1, 2009

### LowlyPion

Where did you get v = .72 from? It fell 1.8m

V2 = 2*9.8*1.8m

How fast does it bounce back?

Isn't that going to give you the ΔV that you want to use?

The 2.5 ms* Fmax looks ok though.