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Free-fall question with extras

  1. Jan 9, 2006 #1
    This problem incorporates the main idea of gravitational freefall, but also adds an extra to it, and I get different answers from the back of the book:

    A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2s later. If the speed of sound is 340m/s, how high is the cliff?

    Book answer: 46m.
    My answer: 52m.

    My solution: Because of the lack of information, I simply used variables to substitute.
    Given: x0 = 0, a = -9.8m/s^2, total time = 3.2s

    x = x0 + v0t + .5at^2 = -4.9t^2 => I changed this to 4.9t^2 to reflect a change in point of reference.

    Since the speed of sound is given, and x is assigned as above, I determined that the time in which sound travelled up the cliff is 340/x sec. Since the problem is asking for distance, I swapped the variables around with the length to get the following: t^2 = x/4.9 => (x/4.9)^.5 => time it takes to fall down the cliff.

    Given that 3.2 s is the total time for the rock to fall AND for the sound to rebound, I wrote down the following: 3.2 = (340/x) + (x/4.9)^.5 . One calculation led to another, and I got the answer at top. Am I right this far and I got my variable switching wrong, or did I do something fundamentally wrong? Eagle eyes most apprecated :D
  2. jcsd
  3. Jan 9, 2006 #2


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    The time for sound to travel x metres at 340 m/s is,

    t = x[m]/340[ms^-1]
    t = x/340
  4. Jan 9, 2006 #3

    Can you explain how you were able to determine that? I don't quite understand.
  5. Jan 9, 2006 #4
    You are on the right track.

    let the distance of the fall = d.

    so d = 340(3.2-t) = 1/2(9.8)t^2

    Use the quad. eq. to solve for t

    ta da
    t = 3.06s, so d = 45.93m
  6. Jan 9, 2006 #5
    So it WAS my calculation! Damn, you made it look so simple! Thanks!!
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