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Rock and Window

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A rock is dropped off the top of a building. On the way down, the rock passes a window. The window is known to be 2.00m tall and the stone takes 0.164s to fall past the window.

    1)What is the velocity of the rock at the top of the window?
    2)What is the velocity of the rock at the bottom of the window?
    3)How much time was necessary from the instant the rock was dropped until it reached the top of the window?
    4)How far above the top of the window was the rock dropped?

    I plugged this but wasn't sure if I started it right.
    xo=0 vox= 0 ax=-9.81m/s^2 t=0.164s x=2.00m Vx=?

    2. Relevant equations
    2a(x-xo)=vx^2-vox^2
    2a(x)=-vox^2
    sqrt 2(9.81m/s^2)(2.00m) = 6.26 m/s? not sure if this is the answer for 1). But I pretty sure it can't be right since question 1 and 2 is asking for top velocity and bottom velocity, which is why I'm confused on how to solve this.


    3. The attempt at a solution
    Attempted once but got stuck on how to continue solving the rest.
     
  2. jcsd
  3. Jan 27, 2013 #2

    tiny-tim

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    Welcome to PF!

    Hi Reth0407! Welcome to PF! :smile:

    Try s = ut + 1/2 at2 :wink:
     
  4. Jan 27, 2013 #3
    Hi tiny-tim! Ok that equation would make sense for half. So velocity of rock at bottom would be the full right which is 6.26m/s?
     
  5. Jan 28, 2013 #4

    tiny-tim

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    6.26 ? :confused:

    oh, that's from …​
    no, both those vs are non-zero, why have you left one out?

    this is the wrong equation (it doesn't help, because you have two unknowns in it)

    you need a constant acceleration equation with only one unknown in it, ie x - xo = vot + 1/2 at2
     
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