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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:
    [tex]H=\frac{(\sqrt{V^{2}+2gVT}-V)^{2}}{2g}[/tex]


    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].

    [tex]T=t_1+t_2[/tex]

    For the falling rock:
    [tex]y(t)=H-\frac{g}{2}t^{2}[/tex]
    [tex]y(t_1)=0[/tex]
    [tex]\frac{g}{2}t_1^{2}=H[/tex]
    [tex]t_1=\sqrt{\frac{2H}{g}}[/tex]

    For the wave of sound:
    [tex]y(t)=Vt[/tex]
    [tex]y(t_2)=H[/tex]
    [tex]Vt_2=H[/tex]
    [tex]t_2=\frac{H}{V}[/tex]

    [tex]T=\sqrt{\frac{2H}{g}}+\frac{H}{V}[/tex]

    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.

    [tex]T-\frac{H}{V}=\sqrt{\frac{2H}{g}}[/tex]
    [tex]T^{2}-\frac{2TH}{V}+\frac{H^{2}}{V^{2}}=\frac{2H}{g}[/tex]
    [tex]\frac{T^{2}V^{2}g}{V^{2}g}-\frac{2TVgH}{V^{2}g}+\frac{gH^{2}}{gV^{2}}=\frac{2V^{2}H}{gV^{2}}[/tex]

    [tex]T^{2}V^{2}g-2TVgH+gH^{2}=2V^{2}H[/tex]

    [tex]gH^{2}-(2TVg+2V^{2})H+T^{2}V^{2}g=0[/tex]
    [tex]H=\frac{2TVg+2V^{2}+\sqrt{(2TVg+2V^{2})^{2}-4T^{2}V^{2}g^{2}}}{2g}[/tex]

    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

    LowlyPion

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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

    diazona

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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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