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 vy1=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, tr, and the time that the sound travels from position 2 to position 1, ts.
\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 tr, 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.