# Falling Objects| Kinematics| Rock Dropped into a Dark Well| College Physics| Problem 2.55

### Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this 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.