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

Adding equation 2 and 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. 

 

 

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