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

[RoyalCat]

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.

 

[LowlyPion]

If you set a pseudo-variable

x = H^1/2

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.

 

[diazona]

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.