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Find initial height when velocity is given

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A flowerpot falls from the ledge of an apartment building. It takes .2 s for the pot to pass the 4 m window below. How far above the top of the window is the ledge from which the pot fell? (Neglect air resistance)

    2. Relevant equations

    y(t)=y0 + (voy)(t) - 1/2g(t^2)
    vy(t) = voy - gt

    3. The attempt at a solution
    4m/.2s =20m/s
    so vy(.2)=voy-(9.81*.2)
    20+1.962= voy
    22= initial velocity

    y(t)=y0 + 22(.2)-(9.81/2)(.2^2)
    0 = y0 +4.4 - .2

    Yeah, I know I'm wrong I just can't figure out how to do this problem.
  2. jcsd
  3. Sep 15, 2011 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    You need to think about the pot's initial velocity again. What happens when you drop something? How fast it is going the moment you let go of it?
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