# How deep is the well, given this data?

A rock is dropped into a well. A timer starts when the rock is dropped and is stopped when the noise of the rock hitting the ground of the well is heard. How deep is the well?

Here's what I have so far: $t = \left(\left(\frac{2\times h}{g}\right)^{\frac{1}{2}}+\left(\frac{h}{340}\right)\right)$ when I try to use the quadratic formula, I can express $h$, but when substitute different amounts for $t$, it doesn't seem right. I get a very small and a very large value for $h$. For example for $t=30\:\mathrm{s}$ I get $3\:\mathrm{m}$ and $42\:\mathrm{km}$, which is quite odd. I also tried it with the following set of equations:

$$\frac{1}{2}\left(g\times t_1^2\right)=h$$ $$t_2\times 340 \:\mathrm{m/s}=h$$ $$t_1+t_2=t$$

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Possible duplicate: physics.stackexchange.com/a/18140/2451 –  Qmechanic Feb 20 '13 at 19:57
If you're squaring the equation $t=\sqrt{2h/g}+h/c$ to get $h$ in terms of $t$, beware of extraneous solutions! Does $h=42\text{ km}$ still satisfy your original equation? –  Emilio Pisanty Feb 20 '13 at 21:38

Let $v$ denote the speed of sound, and let's use your equations. Combine the second and third equations to get $$\frac{h}{v} = t-t_1$$ so that $$t_1 = t-\frac{h}{v}$$ and therefore using the first equation $$h = \frac{1}{2}g\left(t-\frac{h}{v}\right)^2$$ Putting this in standard form to use the quadratic equation gives $$h^2 - 2v\left(t+\frac{v}{g}\right)h + (vt)^2 = 0$$ which gives $$h = \frac{gtv+v^2\pm v\sqrt{v(2gt+v)}}{g}$$ for $g=9.8\,\mathrm{m}/\mathrm{s}^2$, $t=30\,\mathrm{s}$, adn $v=340\,\mathrm{m}/\mathrm s$, I get $2.5\,\mathrm {km}$ and $41.5\,\mathrm{km}$. The first solution is the correct one, for a 30 second fall, I think it seems quite reasonable; if you were freely falling for that amount of time without air resistance, you would travel about 4.4 kilometers.
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