# College Physics 2.55 – A rock dropped into a dark well

## Solution:

### Part A

So, we just need to solve for the distance y using the equation

$y=v_ot+\frac{1}{2}at^2$

Substitute the given values

$y=0+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(2.0\:s\right)^2=-19.6\:m$

Neglecting the speed of sound, the distance to the water is 19.6 m below the ground.

### Part B

Let $h$ be the depth of the well, $y_o=0\:m,\:y=-h$ at the bottom of the well. Let $v_2$ be the speed of sound and $t_2$ be the time for the sound to travel from the bottom to the top of the well. Let $t_1$ be the time for the rock to reach the bottom.

$t_1+t_2=2.0\:s$                                         Equation1

$-h=\frac{1}{2}a\left(t_1\right)^2$                          Equation2

$h=v_2t_2=v_2\left(2.0-t_1\right)$                         Equation 3

$0=\frac{1}{2}a\left(t_1\right)^2+v_2\left(2.0-t_1\right)$

Substitute the given values

$0=0.5\left(-9.8\right)\left(t_1\right)^2+\left(332\right)\left(2-t_1\right)$

$0=-4.9\left(t_1\right)^2+664-332t_1$

Rearranging, we have

$-4.9\left(t_1\right)^2-332t_1\:+664=0$

Solve the equation using the quadratic formula

$t_1=\frac{-\left(-332\right)\pm \sqrt{\left(-332\right)^2-4\left(-4.90\right)\left(664\right)}}{2\left(-4.9\right)}=1.944\:s$

We have solved for $t_1$. Now, we can substitute this value to equation 2 to solve for $h$

$h=-\frac{1}{2}\left(-9.80\right)\left(1.944\right)^2=18.5\:m$

With the speed of sound taken into account, the distance to the water is about 18.5 meters.