# Determine depth of well using speed of sound

#### Verify360

Q: Screenshot
t = 2.5 for sound of stone to reach the top of the well from the bottom
s = (340)(2.5) + 1/2(-9.8)(2.5)^2 = 819.4m
This is wrong according to the textbook, is any other way to approach the problem?

#### skeeter

Math Team

time to drop $\ne$ sound return time

Let $t$ = time for the stone to hit the water
$5-t$ = time for the sound to return

$\dfrac{1}{2}gt^2 = 340(5-t)$

solve for $t$, then evaluate the well depth

#### Verify360

View attachment 10804

time to drop $\ne$ sound return time

Let $t$ = time for the stone to hit the water
$5-t$ = time for the sound to return

$\dfrac{1}{2}gt^2 = 340(5-t)$

solve for $t$, then evaluate the well depth
How did you simplify the RHS to 340(5 - t)? Shouldn't it be 350(5 - t) + 1/2(-9.8)(5 - t)^2?

#### skeeter

Math Team
As stated in my intial post, $t$ is the time required for the stone to drop to the bottom of the well. The problem statement says it takes 5 seconds to hear the splash after the stone's release, that means it takes $(5-t)$ seconds for the sound of the splash to reach the top of the well.

Shouldn't it be 350(5 - t) + 1/2(-9.8)(5 - t)^2?
Sorry, but no. The equation I wrote is ...

distance the stone falls = distance the sound travels up

The falling stone's speed accelerates due to gravity, hence the distance it falls down is $d = \dfrac{1}{2}gt^2$

The speed of sound is a constant, and does not depend on gravity, hence the distance sound travels back up is $d = 340(5-t)$

Common sense should tell you $t > (5-t)$ because the stone falls at a much slower average speed than the speed of sound.

You should get the well's depth to be a value between 100 and 110 meters.

Verify360