# Thread: a falling stone with sound

1. ## a falling stone with sound

You drop a rock into a well (from ground level). You can't see the impact at the bottom but you hear it after 6.1 seconds. You wonder how deep the well is. Assume the speed of sound is 1100 feet per second and the acceleration of the rock is 32 feet per second squared. Ignore air drag (this is a major simplifying assumption).

(Hint: the time till you hear the impact is the sum of the time the rock takes to reach the bottom and the time the sound of the impact takes to travel back to your ear.)

The depth of the well (in feet) is: ????

I don't even know where to start!

2. Originally Posted by MathNeedy18
You drop a rock into a well (from ground level). You can't see the impact at the bottom but you hear it after 6.1 seconds. You wonder how deep the well is. Assume the speed of sound is 1100 feet per second and the acceleration of the rock is 32 feet per second squared. Ignore air drag (this is a major simplifying assumption).
$\mbox{time until bottom }+\mbox{ time until top} = 6.1$.

Now "time until bottom" means the amount of time the rock hits the water. But when it hits the water it takes time for the sound to travel up. That is what I mean by "time until top". In total, this trip takes 6.1 seconds.

Let $s$ be the length of the well.

That means "time until bottom" is $\frac{\sqrt{s}}{4}$. (Why is that. This is an important part of this solution).

Now "time until top" is simply the distance formula. Thus, $\frac{s}{1100}$.

Thus, we have an equation,
$\boxed{ \frac{\sqrt{s}}{4}+ \frac{s}{1100} = 6.1 }$.

Now solve it.