a ball being dropped and bouncing back up?

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.

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.

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?

mgb_phys

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

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?

mgb_phys

Correct method - but be careful of the signs.

needlottahelp

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