# Determine depth of a well using the speed of sound

1. Nov 11, 2013

### slurik

1. The problem statement, all variables and given/known data
If you drop a stone into a deep well and hear a splash 4.68s after dropping the stone, how far down is the water level? Neglect air resistance and assume that the speed of the sound in air is 3.40x10^2 m/s

2. Relevant equations
v=d/t
d=v1t+1/2at^2

3. The attempt at a solution

Here is what I had a shot at:

for sound:
v=d/t
d=vt
d=340t

for the stone:
d=v1t+1/2at^2
d=0(4.86s)+1/2(-9.8m/s^2)(4.86-t)^2

for the system:
d=d
so:
340t=1/2(-9.8m/s^2)(4.68-t)^2
340t=-4.9m/s^2 (t^2-9.36t+21.9024)
340t=-4.9t^2+45.864t-107.32176
0=-4.9t^2-294.136t-107.32176

Using that quadratic, I substitute it into the quadratic formula and obtain 2 extraneous solutions:
t=-59.6606....s or t=-0.367s

I can see that the amount of time it will take the sound to reach the observers ear is very small in comparison to the amount of time it will take the rock to fall to the bottom. as such, I expect t to be very small since sound is travelling at 340m/s. also I noticed that 1/2at^2 is = approx 107m. The best result I have come up with is 124m, which exceeds the above mentioned result.

The solution sheet says 95.0m

Last edited: Nov 11, 2013
2. Nov 11, 2013

### vela

Staff Emeritus
You have a sign mistake. The lefthand side is positive when t>0 while the righthand side is negative because you used a=-9.8 m/s^2.

3. Nov 11, 2013

### slurik

Gah! Thanks so much. Sometimes signs get the better of me when ive been working a long time. I have found the solution. Many thanks again! Time for a break perhaps.