# Pebble being dropped into a well

1. Jun 5, 2012

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

A pebble is dropped into a deep well, and 3.0 seconds later the sound of a splash is heard as the pebble reaches the bottom of the well. The speed of sound in air is about 340 m/s.
(a) How long does it take for the pebble to hit the water?
(b) How long does it take for the sound to reach the observer?
(c) What is the depth of the well?

From College Physics by Serway and Faughn

2. Relevant equations

$v = v_0 + at$

$x = v_0t + \frac{1}{2}at^2$

$v^2 = v_0^2 + 2ax$

3. The attempt at a solution

(a) So in this problem we know the acceleration (which is the gravitational constant) and we also know $v_0$ which is 0 because it starts out as 0 m/s. But in all three equations, there are 4 variables, and knowing 2 variables is not enough.

Then I tried plugging in and obtaining a system. So I got

$-19.6x = v^2$
$v = -9.8t$

But this doesn't help much, and if you add a 3rd equation its going to be the same thing (the three equations are related). So I don't know how to proceed :(

Also for (b) and (c) I think you need (a) to solve them.

BTW I'm out of school, so this problem doesn't count for a grade or anything :)

2. Jun 5, 2012

### NemoReally

Forget the distance for the moment and try to find the time. You need the speed of sound for this.

3. Jun 5, 2012

### SammyS

Staff Emeritus
How have you defined your variables?

I don't see any of your solution which includes the speed of sound.

Don't assume that you can get part (a) without also working on (b) and/or (c) .

4. Jun 5, 2012

### HallsofIvy

Staff Emeritus
Taking the depth to be d, the time it takes for the rock to hit the water is $t_1= \sqrt{d/4.9}$ and the time for the sound to come back is $t_2= d/340$. So the total time is given by $t_1+ t_2= \sqrt{d/4.9}+ d/340= 3$.

5. Jun 5, 2012

@ SammyS: Oops yeah, I did forget to define variables. In this case $v_0$ is the starting velocity, $t$ is the time which it takes for the pebble to hit the water, and $x$ is the displacement. (in this case it's also the distance because the pebble doesn't go back) And $v$ is the velocity when the pebble hits the water, though I guess that wasn't needed.