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Drowsy Cat question

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A drowsy cat spots a flower pot that sails first up the window, and then down past an open window. The pot is in view for a total of 0.5s, and top to bottom height of the window is 2 m. How high above the window top does the flower pot go?

    2. Relevant equations
    Using kinematics equations:
    h = v.t + 0.5*a*t^2
    v[f]-v = at
    or, 2h/t = v[f]+v

    3. The attempt at a solution
    I assumed that it is going down, while it was viewed.
    so, here's my diagram:

    ^2 m, 0.5 s

    (motion is opposite to the direction indicated by the carrot symbols)

    so using those equations I get
    v = 1.5475 m/s

    and as 2as = v^2

    so s = 0.122 m

    but the answer at the back of the book is 2.34 m ><
  2. jcsd
  3. Sep 22, 2007 #2
    Last edited: Sep 22, 2007
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