AP Physics: Solving Average Acceleration Problem

AI Thread Summary
The discussion revolves around solving an average acceleration problem involving a tennis ball dropped from a height of 4.00 m that rebounds to 2.00 m. The user initially calculates the impact velocity using the equation v_f^2 = v_0^2 + 2aΔy, finding it to be -8.85 m/s. They struggle to determine the rebound velocity (v2) needed for the ball to reach 2.00 m, which is crucial for calculating average acceleration. After realizing that v2 must be positive as the ball moves upward, they successfully compute it and find the correct average acceleration. The interaction highlights the importance of understanding both downward and upward motion in physics problems.
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Hey everyone,

I'm taking AP Physics B and self-studying AP Physics C this year. School hasn't started yet, but I am already stuck on this problem from Fundamentals of Physics by Halliday, Resnick, & Walker.

Homework Statement



A tennis ball is dropped onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, what is its average acceleration during that contact?

v0=0 m/s, y0=4.00 m, a=-9.8 m/s2

Homework Equations



\Delta y = \frac{1}{2}at^2+v_0t

v_f^2=v_0^2+2a\Delta y

The Attempt at a Solution



I used v_f^2=v_0^2+2a\Delta y to find the velocity when the ball hits the ground.

v_f=-\sqrt{v_0^2+2a\Delta y}=-\sqrt{0^2+2*-9.8*-4} = -8.85 m/s

I know that average acceleration is calculated by a_{avg}=\frac{v_2-v_1}{t_2-t_1}

I just calculated v1 to be -8.85 m/s and the problem gives that t2-t1 is equal to 0.012 s.

This is where I got lost. I'm not sure how to calculate v2. I think it should be positive because during the contact, the ball changes from moving downwards to upwards.

Can anyone help?
 
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Don't forget that the ball rebounds to a height of 2.0 m. At what speed would it have to leave the floor, heading upward, in order to come to rest at that height? That velocity is your v2 and, since you called "downward" negative, this velocity will have a positive sign, making your average acceleration positive as well (that is, "upward").
 
Wow! I think my mind must have decided that piece of information was insignificant and automatically disregarded it each time I read it. :redface:

Thanks for your help! I was able to calculate v2 and get the correct answer! :smile:
 
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