AP Physics: Solving Average Acceleration Problem

[azure kitsune]

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?

 

[dynamicsolo]

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").

 

[azure kitsune]

Wow! I think my mind must have decided that piece of information was insignificant and automatically disregarded it each time I read it.

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