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Average Acceleration

  1. Aug 21, 2008 #1
    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.

    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    [tex]\Delta y = \frac{1}{2}at^2+v_0t[/tex]

    [tex]v_f^2=v_0^2+2a\Delta y[/tex]

    3. The attempt at a solution

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

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

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

    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?
     
  2. jcsd
  3. Aug 21, 2008 #2

    dynamicsolo

    User Avatar
    Homework Helper

    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").
     
  4. Aug 22, 2008 #3
    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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