# A ball being dropped and bouncing back up?

1. Oct 5, 2008

### needlottahelp

1. The problem statement, all variables and given/known data
After falling from rest at a height of 30m, a 0.50 kg ball rebounds upward, reaching a height of 20m. If the contact between ball and ground lasted 2.0 ms, what a average force was exerted on the ball?

2. Relevant equations
The kinematic equations?
F=ma
w=mg?
I dont really know

3. The attempt at a solution
Sorry I don't even know where to begin. I don't really understand this section in the book so i dont even know what to do. The answer is supposed to come out to 1.1 x 10^4 upward.

Last edited: Oct 5, 2008
2. Oct 5, 2008

### mgb_phys

You need to work out the speed the balls hits the floor and then the speed it starts backup at. Acceleration is just change in speed / time.

Hint - the speed it starts back up is the same as the speed a ball dropped from the second height would reach the floor.

3. Oct 5, 2008

### needlottahelp

Ok so the speed of the ball when it hits the ground first time would need the kinematic equation Vf^2= Vi^2 + 2ad right?

Vf^2= (0)^2 + 2(9.8)(30)
Vf^2 = 588
Vf~24.3 m/s ?

Vf^2= 2(9.8)(20)
Vf^2 = 392
Vf ~ 19.8 m/s ?

I think thats correct. correct me if I'm wrong please. How do I find the acceleration?

4. Oct 5, 2008

### mgb_phys

Acceleration = change in velocity / time
Whats the total change in velocity (watch the signs!)

5. Oct 5, 2008

### needlottahelp

Ohhh...

So the acceleration is the change in velocity which would be 19.8 - 24.3/change in time.
Then put it in the F=ma and solve for F correct? or is there more?

6. Oct 5, 2008

### mgb_phys

Correct method - but be careful of the signs.

7. Oct 5, 2008

### needlottahelp

OK thank you very much. I was stuck on this problem for a while.