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How high is the cliff?

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    To find the height of a cliff, you drop a rock from the top and 10s ater, you hear the sound of it hitting the ground at the foot of the cliff. Ignoring air resistance, how high is the cliff if the speed of sound is 330m/s.

    2. Relevant equations

    Xf = Xo = VoT + 1/2aT^2
    Vf = Vo + aT
    Vf^2 = Vo^2 + 2a(deltaX)

    V = (deltaX)/(deltaT) maybe?

    3. The attempt at a solution

    ok so here is what i tried.
    i drew it so that
    T1 = time for rock to fall
    T2 = time for sound to go back
    Tt = Total time = 10s
    so
    T1+T2 = Tt = 10s

    then I did
    Vs = velocity of sound, so
    Vs = deltaX / deltaT
    solved for deltaX which = height of the cliff and got

    deltaX = Vs(Tt-T1)

    then for the rock,
    Vf^2 = Vo^2 +2a(deltaX)
    knowing that Vo^2 = 0, and a = g, and solving for deltaX
    deltaX = Vf^2/2g

    since both equations are solving for the same deltaX, I set them equal to eachother
    Vs(Tt-T1) = Vf^2/2g

    the unknowns are T1, and Vf

    I feel pretty good up to this point, then i'm not so sure, but next i tried...
    to solve for T2 by
    Vf = Vo +at
    t is for the rock so...
    Vf = g(Tt-T2)

    then plug that in and get the equation

    Vs(Tt-T1) = g(Tt-T2)^2/2g
    but there's still two unknowns with T1 and T2, so...
    if (Tt-T2) = T1, can i plug that in on the right side, so that T1 is the only variable? it's just on both sides?

    am i close? where did i mess up?
     
  2. jcsd
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