Bouncing ball average acceleration

AI Thread Summary
The discussion revolves around calculating the average acceleration of a golf ball that bounces after being dropped from a height of 1.5 meters. The ball rebounds to a height of 1.1 meters, and the contact time with the floor is 6.2 x 10^-4 seconds. Participants are clarifying the correct use of variables, including initial and final velocities, and the application of relevant equations for motion under gravity. There is confusion regarding the calculation of velocities and the correct interpretation of the time of contact. The thread emphasizes the importance of understanding the direction of velocities and accurately applying the formulas to find the average acceleration.
physicsquest
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Homework Statement


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?


Homework Equations


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


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?
 
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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.
 
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.
 
physicsquest said:
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.
Refer the post " dropping a tennis ball " by demonelite.
 
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