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Free-Fall Motion (w/ a tennis ball)

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00m. It rebounds to a height of 2.00m. If the ball is in contact with the floor for 12.0ms, (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down?


    2. Relevant equations
    Equations of motion w/ constant acceleration


    3. The attempt at a solution
    I'm pretty sure (b) is up for starters since the ball reaches a height of 2m after it is on the ground which means it's got to accelerate up.

    My problem is with part a. I started by figuring out the velocity of the ball the moment it hits the ground.

    [tex]v = at[/tex]

    [tex]v = \frac{y-y_o}{t}[/tex]

    So substituting the second equation in the first one and solving for t:
    [tex]t = \sqrt{\frac{y-y_o}{a}}[/tex]

    [tex]t = \sqrt{\frac{-4m}{-9.80\frac{m}{s^s}}} = .64s[/tex]

    Since I got the time...

    [tex]v = v_{o}+at = 0 + (-9.80)(0.64) = -6.3m/s[/tex]

    In my mind I thought that I had the initial velocity of the time interval it was on the ground and I thought that the final velocity would be 0 since if it was greater than 0 the ball would be traveling "up".

    First I found the time in s and got 12.0ms = .012s, then:
    [tex]a_{avg} = \frac{v-v_o}{t} = \frac{0-(-6.3)}{.012}[/tex]

    My answer was a = 525m/s^2, but it's wrong -- my book says its 1.26*10^3 m/s^2

    So uhh... where did I go wrong? XD
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 9, 2007 #2
    Oh don't think it would really matter at all in this problem but assume there is no air resistance.
     
  4. Sep 9, 2007 #3

    bel

    User Avatar

    How did you get the time? Try using [tex]s=ut+\frac{1}{2}at^2[/tex] and [tex]s=vt-\frac{1}{2}at^2[/tex]. Then applying [tex] v=\frac{y_1-y_o}{t}[/tex], find the velocity just before it hits and just after it rebounds, use [tex] a=\frac{(v-u)}{t_o}[/tex], remembering that one velocity vector points downward and the other vector points upward.
     
  5. Sep 9, 2007 #4
    Whoops should have made it a bit clearer how I derived the time. I edited my first post.

    But I really don't get what your saying to do...
     
  6. Sep 9, 2007 #5

    bel

    User Avatar

    Well, get two different veocities, by the first three equations, and use the fourth to find your acceleration.
     
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