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Flower pot falls past a window

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is L. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

    Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

    From what height h above the bottom of your window was the flower pot dropped?


    2. Relevant equations

    So here's what I did...

    I set L=Viw*t+.5gt^2, where Viw is equal to the initial velocity of the flower pot at the top of the window.

    Then I set Viw=Vfd=sqrt(2gx), where Vfd is the velocity of the pot after it reaches the top of the window and x is equal to the displacement from the dropping point to the top of the window.

    3. The attempt at a solution

    So here is what I tried,

    First I solved for Viw

    Viw=(L-.5gt^2)/(t)

    Then I set this equal to the quantity Vfd or...

    Vfd=Viw=(L-.5gt^2)/(t)=sqrt(2gx)

    Finally I solved for x+L and got

    (4L^2-4Lgt^2+g^2t^4+8gLt^2)/(8gt^2)=x+L=h

    I know my answers wrong, but I'm not sure why? I've rechecked my work several times and there's something that I'm not seeing. Can somebody please help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 17, 2010 #2
    Here's what I would say

    Up until the window
    x=.5gT^2
    Vw=gT
    T=Vw/g

    through the window
    L=Vw*t+.5g*t^2
    Vw=(L-.5gt^2)/t

    all together
    H=x+L
    H=Vw^2/(2g)+L
    H=(L-.5gt^2)^2/(2g*t^2)+L

    whatever that works out to be :)
     
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