How Does a Golf Ball Accelerate Upon Impact with Concrete?

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A golf ball released from a height of 1.76 m above a concrete floor, bounces back to a height of 0.80 m. If the ball is in contact with the floor for 4.62 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?

I'm not sure how to start this. I was thinking of this equation:

4.7x^2-1.76x

I think
 
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chaotixmonjuish said:
A golf ball released from a height of 1.76 m above a concrete floor, bounces back to a height of 0.80 m. If the ball is in contact with the floor for 4.62 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?

I'm not sure how to start this. I was thinking of this equation:

4.7x^2-1.76x

I think
Why?

What is the velocity of the ball when it hits the floor?

What is the initial velocity of the ball in order to achieve an altitude of 0.80 m?

What is the change in velocity?

What is the average acceleration thereof?
 
I'm not sure, I lifted this question off my homework. I'm pretty stumped on this question.
 
chaotixmonjuish said:
I'm not sure, I lifted this question off my homework. I'm pretty stumped on this question.

What is the velocity right before it hits the ground? What is the velocity right immediately after it bounces?