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Homework Help: HELP! Tough Question

  1. Jan 9, 2006 #1
    Ive been working on this and its a little tough: Whats the answer?

    A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.20s for the tile to pass her window whose size is 1.6m. How far above the top of this window is the roof?
    (Hint: 2 calculations- Use uniform acceleration equations)

    How am i suppose to do this?

    a=9.81m/s^2 (im in canada we use 9.81 not 9.8)
    t= 0.20s
    d= ?
    Vi= 0m/s

  2. jcsd
  3. Jan 9, 2006 #2


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    Hint: the initial velocity (that is, the velocity as it passes the top edge of the window) is not zero. You need to solve for this velocity, as it will tell you how far from the roof the top of the window is.

    - Warren
  4. Jan 9, 2006 #3
    but the rooftile falls from rest rest= 0m/s
  5. Jan 9, 2006 #4
    is the answer 2.51 meters?
  6. Jan 9, 2006 #5


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    The roof tile is NOT AT REST when the observer first observes it.

    First, it falls from the roof to the top of the window, gaining some velocity. Next, it falls past the window, which the observer times at 0.2 seconds.

    Its velocity, when the observer first sees it, is NOT zero. You need to solve for this velocity first. You can then use it to solve for the height of the roof above the window.

    - Warren
  7. Jan 9, 2006 #6
    Heres the caluclations

    s = 1.6 t= .2 a = g = 9.81 v =?
    s = vt +at^2
    and v turns out to be 6.038
    well this is the velocity when it reached the point -lets say- x which is when observer first sees it. because it fell off the roof and travelled lets say distance d already...so...
    d = (v^.5)/2g
    = 1.86 m

    i think thats pretty much it.
  8. Jan 10, 2006 #7


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    s = (1/2)gt^2


    s+1.6 =(1/2)g(t+0.20)^2

    think!! and eliminate t.

  9. Jan 10, 2006 #8
    Not going to give an answer here but the working I would use seems to differ slightly from others, and arrives at a slightly different expression.

    we know that distance = 1/2(inital velocity + final velocity) x time or...
    s =1/2(u+v)t
    we also know that velocity (v) = u + at
    using substitution you can end up with an equation that involves only one variable...that is 'u'
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