Depth of a Pit Based on Time Delay of Sound from Falling Rock

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The discussion revolves around calculating the depth of a pit based on the time delay of sound from a falling rock. The relationship between the total time T, the time taken for the rock to fall (t_1), and the time for sound to travel back up (t_2) is established. The equations derived show that T can be expressed in terms of the depth H, the speed of sound V, and gravitational acceleration g. Participants confirm that the algebraic manipulation leading to the depth formula H = ((√(V² + 2gVT) - V)²) / (2g) is correct, with suggestions for simplifying the calculations. The thread emphasizes the importance of careful algebraic steps to arrive at the solution.
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Homework Statement



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


Homework Equations


Kinematic equations.


The Attempt at a Solution


T is the time it takes for two motions to transpire. The first is the stone falling, t_1. The second, is the sound traveling back up, t_2.

T=t_1+t_2

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

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

T=\sqrt{\frac{2H}{g}}+\frac{H}{V}

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.

T-\frac{H}{V}=\sqrt{\frac{2H}{g}}
T^{2}-\frac{2TH}{V}+\frac{H^{2}}{V^{2}}=\frac{2H}{g}
\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}}

T^{2}V^{2}g-2TVgH+gH^{2}=2V^{2}H

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

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 V^{2} looks like it's going to be a problem.
 
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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.
 
No you haven't made a mistake, and you've got just a few small algebraic steps left. Try expanding the (2TVg + 2V^2)^2 under the square root.

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