# Depth of a Water Well by dropping a rock

1. Feb 4, 2014

### MienTommy

1. The problem statement, all variables and given/known data

A man drops a rock into a well. He hears the sound of the splash 3.2 seconds after he releases the rock from rest. The speed of sound in air (at the local ambient condition) is 338 m/s

(a) How far below the top of the well is the surface of the water? (round your answer to a whole number)

(b) If you ignored the travel time for the sound, what would have been the calculated depth? (round your answer to a whole number)

time of rock + time of sound = 3.2s
acceleration = -9.8 m/s^2
V_i = 0 m/s
V_sound = 338 m/s

2. Relevant equations

x=x_0 + v_0*t + (1/2)at^2

3. The attempt at a solution

x = 0 + 0 + (1/2)(-9.8)(3.2)^2

However, I realized, this is incorrect because I also need to account the time the sound took to travel to the person's ear subtracted by the 3.2 seconds. My problem is, how do I find the time it took to reach the person's ear when the rock hit the surface of the water?

2. Feb 5, 2014

### voko

Obviously, the rock takes time $t_1$ to hit the water, and the sound takes time $t_2$ to reach you. As you wrote, $t_1 + t_2 = t = 3.2 \ \text{s}$.

3. Feb 5, 2014

### PeroK

You want to find the depth of the well, d. So, try to derive an equation that relates t and d.

For example, if d = 10m, it would be quite easy to work out t. So, try to use this approach to relate d and t generally.

4. Feb 5, 2014

### Abaidullah

Hi
There r also some different issues like drag force of water on shape of rock so how this is possible