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Falling Object

  1. Feb 12, 2004 #1
    Hey

    This seems to be way easy, but i'm just not gettng where to start it. A stone is dropped from the top of a cliff, it is seen to hit the ground below after 3.66 s, how high is the cliff? Anyone know? Thanks.
     
  2. jcsd
  3. Feb 12, 2004 #2

    Tom Mattson

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    That means that vi=0, and the yi the height of the cliff, which is unknown. Implicit in this is that the acceleration of the stone is that of gravity.

    That means that Δt=3.66s. Implicit in that is that yf=0, which is ground level.

    Can you find an equation that relates those quantities?
     
  4. Feb 12, 2004 #3
    Mentor Edit: Please don't post complete solutions. Thank you.
     
    Last edited by a moderator: Feb 12, 2004
  5. Feb 12, 2004 #4
    Tom-

    I pretty much got that far on my own. I just don't know which formula ot use. The one that looks the closes would be

    /\=Delta

    /\X=X-Xo=VoxT+1/2AxT^2

    and

    Vx^2-Vox^2=2Ax/\x

    but i'm not sure. I'm having a really hard time figuring out which formula goes where. Thanks.

    -Alice
     
  6. Feb 12, 2004 #5

    turin

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    This one always works for a constant acceleration. And, of course, that "X" doesn't have to be horizontal.
     
  7. Feb 13, 2004 #6

    Tom Mattson

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    Right, the equation also holds for "y".

    Alice, look at the equation that turin referred to, rewrite it for y (just replace x with y), and verify that you do in fact have all the information to solve the problem. Then, plug in the numbers.
     
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