College Physics by Openstax Chapter 2 Problem 55: Determining the Depth of a Well from the Time Delay of a Returning Sound

Problem:

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

Consider Figure A.

We shall consider two points for our solution. First, position 1 is the top of the well. In this position, we know that y₁=0, t₁=0 and v_y1=0.

Position 2 is located at the top of the water table where the rock will meet the water. Since we neglect the time for the sound to travel from position 2 to position 1, we can say that t₂=2.0000 s, the time of the rock to reach this position.

Solving for the value of y₂ will determine the distance between the two positions.

\begin{align*} \Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2-y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = y_1 +v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 2.0000\ \text{s} \right)^2 \\ y_2 & = -19.6\ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer }\right)} \end{align*}

Position 2 is 19.6 meters measured downward from position 1.

$\therefore$ The distance to the water is about 19.6 meters.

Part B

For this case, the 2.0000 seconds that is given includes the time that the rock travels from position 1 to position 2, t_r, and the time that the sound travels from position 2 to position 1, t_s.

\begin{align*} t_r+t_s & =2.0000\ \text{s} \\ t_s & = 2.0000\ \text{s}-t_r \end{align*}

Considering the motion of the rock from position 1 to position 2.

\begin{align*} \Delta y & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2-y_1 & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2 & = y_1 +v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( t_r \right)^2 \\ y_2 & = -4.905\left( t_r \right)^2 \qquad {\color{Blue} \text{Equation 1}}\\ \end{align*}

Now, let us consider the motion of the sound from position 2 to position 1. Sound is assumed to have a constant velocity of 322.00 m/s.

\begin{align*} \Delta y & = v_s \times t_s \\ y_1-y_2 & =\left( 322.00\ \text{m/s} \right)\left( t_s \right) \\ 0-y_2 & =\left( 322.00 \right)\left( 2.0000-t_r \right) \\ y_2 & = -322.00\left( 2.0000-t_r \right) \qquad {\color{Blue} \text{Equation 2}} \end{align*}

So, we have two equations from the two motions. We can solve the equations simultaneously.

\begin{align*} -4.905 \left( t_r \right)^2 & = -322.00\left( 2.0000-t_r \right) \\ 4.905 \left( t_r \right)^2 & =322.00\left( 2.0000-t_r \right) \\ 4.905 \left( t_r \right)^2 & = 644.00-322.00t_r \\ 4.905\left( t_r \right)^2 + 322.00t_r-644.00 & = 0 \\ \end{align*}

We can solve the quadratic formula using the quadratic equation.

\begin{align*} t_r & = \frac{-b \pm\sqrt{b^2-4ac}}{2a} \\ t_r & = \frac{-322.00\pm\sqrt{\left( 322.00 \right)^2-4\left( 4.905 \right)\left( -644.00 \right)}}{2\left( 4.905 \right)}\\ t_r & =1.9425 \ \text{s} \end{align*}

Now that we have solved for the value of t_r, we can use this to solve for y₂ using either Equation 1 or Equation 2. We will use equation 1.

\begin{align*} y_2 & = -4.905\left( t_r \right)^2 \\ y_2 & = -4.905 \left( 1.9425 \right)^2 \\ y_2 & =-18.5 \ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Position 2 is about 18.5 meters below position 1.

$\therefore$ In this case, the distance between the two positions is 18.5 meters.

College Physics Chapter 2 Problems

Problem 1	Problem 2	Problem 3	Problem 4	Problem 5
Problem 6	Problem 7	Problem 8	Problem 9	Problem 10
Problem 11	Problem 12	Problem 13	Problem 14	Problem 15
Problem 16	Problem 17	Problem 18	Problem 19	Problem 20
Problem 21	Problem 22	Problem 23	Problem 24	Problem 25
Problem 26	Problem 27	Problem 28	Problem 29	Problem 30
Problem 31	Problem 32	Problem 33	Problem 34	Problem 35
Problem 36	Problem 37	Problem 38	Problem 39	Problem 40
Problem 41	Problem 42	Problem 43	Problem 44	Problem 45
Problem 46	Problem 47	Problem 48	Problem 49	Problem 50
Problem 51	Problem 52	Problem 53	Problem 54	Problem 55
Problem 56	Problem 57	Problem 58	Problem 59	Problem 60
Problem 61	Problem 62	Problem 63	Problem 64	Problem 65
Problem 66