Physics
posted by Lindsay .
A golf ball released from a height of 1.80 m above a concrete floor, bounces back to a height of 1.06 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?
Quidditch told me that the first steps would be to find the velocity as the ball first makes contact with the floor, and the velocity as the ball leaves the floor. But I'm still unsure how to do this...

the velocity when the ball first contacts the floor is:
V1 = sqrt(2*g*s)
V1= sqrt(2 * 9.8(m/s^s) * 1.8m)
The velocity when the ball just leaves (rebounds) works the same way except for the distance sign
V2 = sqrt(2 * 9.8(m/s^2) * (1.06m))
V1  V2= delta V (you will be subtracting a negative number remember for V2.
It is given that delta t is 4.62ms
acceleration = (delta v) / (delta t)
Don't forget to change the ms time to seconds. 
THANK YOU!
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