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Finding height with constant acceleration

  1. Oct 3, 2008 #1

    !!!

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    1. The problem statement, all variables and given/known data
    What's the cliff's height (m) if a fallling boulder takes 1.3 s to fall the last 3rd of the way to the ground?
    Air resistance is ignored, and the problem requires a quadratic formula.

    2. Relevant equations
    I think it's the constant x-acceleration formulas, which I can't type here because of the super subscripts/ subscripts.


    3. The attempt at a solution
    Actually, I don't even know how to begin this or set this up. =/ How do I solve this equation and determine which one of the given answers is correct? The answers in my textbook says it's either 246 m and 2.51 m, but how do I solve to get them?
     
  2. jcsd
  3. Oct 3, 2008 #2

    cepheid

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    Call the cliff's height h. Let y(t) be the distance above the ground as a function of time.

    [tex] y(t) = h - \frac{1}{2}gt^2 [/tex]

    I'm assuming it starts from rest. Solving for t, the total time to fall is:

    [tex] t = \sqrt{\frac{2h}{g}} [/tex]

    1.3 s before that elapsed time, the ball has fallen 2/3 of the way:

    [tex] y\left(\sqrt{\frac{2h}{g}} - 1.3 s\right) = \frac{h}{3} = h - \frac{1}{2}g\left(\sqrt{\frac{2h}{g}} - 1.3 s\right)^2 [/tex]

    Now I think you can just solve for h. Let me see if I get one of those answers.
     
  4. Oct 3, 2008 #3

    cepheid

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    edit: Yes, I do get both of those answers.
     
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