# Depth of a well (using speed of sound)

1. Nov 26, 2015

### halo168

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

2. Relevant equations
• How does the speed of sound play a role in this?
• How can I find the depth?
3. The attempt at a solution
I used v= [(331)*sqrt(1 + T/273)]. If v = (lambda)/t, then won't lambda equal [(331)*sqrt(1 + T/273)] * t?
I assumed that t would equal 1 s because it takes 1 s for the stone to hit the bottom and 1 s for the echo to be heard (2s total). Is that a correct assumption? As a result, my answer was that lambda = 337.01 m but the correct well depth were supposed to be 18.5 m...

What did I do wrong?

2. Nov 26, 2015

### Staff: Mentor

The stone accelerates as it falls, starting from zero speed. It takes some time to fall to the bottom of the well. The sound it makes when it hits the water travels back up the well at the speed of sound, which is a constant speed and much faster than the stone ever was moving. So the time for the two paths (stone falling, sound rising) is not the same.

3. Nov 26, 2015

### Staff: Mentor

If it took the same time for the stone to hit the water as it took for the sound to reach you, that means that the stone broke the speed of sound!

You need an equation to account for the position of the stone as a function of time.

4. Nov 26, 2015

### halo168

The main problem is that I am not sure why the depth is 18.5 m while I had solved for lambda and got 337.01 m. What did I do wrong in terms of substitution into the equation?

5. Nov 26, 2015

### Staff: Mentor

You don't know how long it took for the sound travel up the well. You have to find at what time the stone hit the bottom.

6. Nov 26, 2015

### azizlwl

You right if you're using echo as distance measurement.

7. Nov 26, 2015

### DaveC426913

Dr Claude has your answer. It is a two part problem. You must solve part 1 first.