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It's a physics-related problem, but it has a nasty equation:

Let the speed of sound be $340\dfrac{m}{s}$, then let a heavy stone fall into the well. How deep is the well when you hear the impact after $2$ seconds?

The formula for the time it takes the stone to fall and the subsequent sound of impact to travel upwards is simple enough:

$t = \sqrt{\dfrac{2s}{g}} + \dfrac{s}{v}$

for $s$ = distance, $g$ = local gravity and $t$ = time.

This translates to said nasty equation:

$gs^2 - 2sv^2 + 2gstv + gt^2v^2 = 0$

Now I need to solve this in terms of $s$, but I'm at a loss as to how to accomplish this. How to proceed?

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    I'm quite ashamed to admit that I kind of forgot to plug in the numbers.2012-06-02

2 Answers 2

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Plugging $g$, $t$ and $v$, your equation will be:

$10s^2 - 2(340)^2s + 4 \cdot 10 \cdot 340s + 10 \cdot2^2 (340)^2 = 0$

Which is a simple quadratic equation. You have permission to solve it now.

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Hint: if you insert the values of $g, t$ and $v$ you have a quadratic equation in $s$. Even if you just regard $g, t$ and $v$ as constants, you can plug this into the quadratic formula.