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Falling roof tile

  1. Jan 30, 2006 #1
    A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.19 s for the tile to pass her window, whose height is 1.4 m. How far above the top of this window is the roof?

    My train of thought is:
    You are given the initial Velocity (Vo=zero m/s), acceleration (-9.8 m/s^2), and the instantaneous velocity of when the tile passes the window (1.4/0.19 = 7.36 m/s)

    So you use the instantaneous velocity and plug it into the equation
    Vf=Vo+(a)(t) to find the amount of time it took to get to the window.

    But if I do that, I get a negative time. I'm not sure where to go here because I can't proceed with a negative time.

    After that, my idea was to take the time, and insert it into another equation and find the distance. Even if I neglect that I get a negative time, I still get the incorrect answer after plugging it in to a different equation. So I think I have two problems I need to overcome.

    Thanks for the help,
  2. jcsd
  3. Jan 30, 2006 #2


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    Homework Helper

    What do you mean by "instantaneous velocity of when the tile passas the window"?
    Velocity at the top of the window is not 0.
  4. Jan 30, 2006 #3
    What I gathered is that the roof tile has an initial velcity of zero, since the problem says it starts from rest.

    The instantaneous velocity is the velocity the woman observes as the tile passes her window. I calculated this by dividing 1.4 meters by 0.19 seconds to get 7.36 m/s

    I figured by finding the velocity of the tile at that moment, you can use that to find out how long it took the tile to reach that velocity. Then using that time it took, you find out the height between the roof top and the window.

    That makes sense to me, and seems like it should work out for me, but it just isn't for some reason.
  5. Jan 30, 2006 #4


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    The trouble is, you need to know at what distance, and at what time did it have that velocity. Since this is an accelerating tile, it requires some work to find that.

    You can say that the top of the window is at x, you know that it had 0 velocity at the roof. Call the roof x=0, so if you call down positive the distance to the top of the window is given by

    [tex] x = \frac g 2 t^2 [/tex]

    now can you use that expression to express the postion and time as it passes the bottom of the window?

    Think about what changes.
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