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Stone thrown down a pit

  1. Jun 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A rock is dropped down a deep pit. The sound of its impact with the floor is heard after T seconds.
    The speed of sound is a given, V.

    Prove that the depth of the pit is:

    2. Relevant equations
    Kinematic equations.

    3. The attempt at a solution
    T is the time it takes for two motions to transpire. The first is the stone falling, [tex]t_1[/tex]. The second, is the sound traveling back up, [tex]t_2[/tex].


    For the falling rock:

    For the wave of sound:


    Now all that remains, so it would seem, is to extract H, since we have one equation with just one variable, since T,V and g are all given.




    And from here, the algebraic manipulation eludes me. Have I made a mistake somewhere along the way? I was thinking about completing the square or some other nonsense, but the coefficient of 2 before [tex]V^{2}[/tex] looks like it's gonna be a problem.
  2. jcsd
  3. Jun 29, 2009 #2


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    Homework Helper

    If you set a pseudo-variable

    x = H1/2

    You can say that you have a quadratic in x ...

    1/V*x2 + √(2/g) * x - T = 0

    Then square the resulting solution.

    I haven't done it, but I'd say a cursory inspection suggests a similar result to where you are.
  4. Jun 29, 2009 #3


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    Homework Helper

    No you haven't made a mistake, and you've got just a few small algebraic steps left. Try expanding the [itex](2TVg + 2V^2)^2[/itex] under the square root.

    If you're still stuck, try working backwards: starting from the solution you're supposed to end up with, expand [itex](\sqrt{V^2 + 2gVT} - V)^2[/itex] and see if you can convince yourself that it's the same as what you got.
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