How Does a Golf Ball Accelerate Upon Impact with Concrete?

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A golf ball dropped from a height of 1.76 m bounces back to 0.80 m after impacting a concrete floor. The contact time with the floor is 4.62 ms, prompting questions about the ball's velocity upon impact and immediately after the bounce. To determine the average acceleration while in contact with the floor, calculations involving initial and final velocities, as well as the change in velocity, are necessary. The discussion highlights the need for equations to analyze the ball's motion and acceleration. Understanding these dynamics is crucial for solving the problem effectively.
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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?
 

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