# Bouncing ball average acceleration

1. Sep 14, 2009

### physicsquest

1. The problem statement, all variables and given/known data
A golf ball released from a height of 1.5m above a concrete floor bounces back to a height of 1.1m. If the ball is in contact with the floor for 6.2 X 10^-4 s, what is the average acceleration of the ball while in contact with the floor?

g= 9.81 m/s^2
x-xo= -1.5
vo= 0
t= 6.2 X 10^-4

Are these variables all correct?

2. Relevant equations
t= ((-2(x-xo))/g)^(-1/2)
average acceleration= (v2-v1)/(t2-t1)

3. The attempt at a solution
t= ((-2(-1.5m)/(9.81m/s^2))^(1/2)= 0.55s

Do I take initial velocity divided by this time number?
And is final velocity just 1.1 divided by time?

2. Sep 14, 2009

### rl.bhat

Using the formula
vf^2 = vi^2 + 2gh, find vf when ball reaches the ground and vi when it rebounds from the ground.
Then find the change in the velocity ( be careful about the directions of the velocities). Time of contact is given. Find the average acceleration.

3. Sep 15, 2009

### physicsquest

I'm still not sure what vf and vi are though. I tried dividing height by time and I used t= (-2(x-xo)/g)^1/2 to find times, but I'm still not getting the right answer.

4. Sep 15, 2009

### rl.bhat

Refer the post " dropping a tennis ball " by demonelite.