# Depth of a well (using a stone and sound)

1. Jan 31, 2009

### needlottahelp

1. The problem statement, all variables and given/known data
A stone is dropped from rest into a well. The sound of the splash is heard exactl 2.00 s later. Find the depth of the well if the air temperatrue is 10.0 degrees Celsius.

2. Relevant equations
Possibly the Kinematics Equation
V= Vo sqrt(1+Temp/273)

3. The attempt at a solution
First i found the speed of the sound and came up with
v = 337.0 m/s

i used the kinematics equation to find the time it took for the stone to hit the bottom of the well and it is

Tstone = sqrt(2x/g)

so i the total time equals the time it takes for the stone to hit the ground and the sound to come up so

2.0 s = x/337 + sqrt( 2x/9.8 ) where x equals the depth of the well

I don't think it is right, and I can't seem to solve for x. Can anyone help? The answer should be 18.5 m

2. Jan 31, 2009

### Delphi51

Looks correct to me! Kind of a messy equation, though.
I used to give that question to my grade 11 physics class so they would have to learn the method of trial and error (guess and test). You can zero in on the right distance in 4 or 5 trials. Too bad you already know the answer.

If you take the x/337 term to the other side so the root is isolated on one side, then square both sides, you will have a quadratic equation. You will no doubt know how to solve a quadratic without resorting to trial and error.

3. Jan 31, 2009

### davieddy

Looks OK to me.