Question

1. A man drops a rock into a well. (a) The man hears the sound of the splash 2.40 s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water? (b) What If? If the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?

Answer 1

a) Let's write the time that the rock takes to heat the water:

$$t=2.4\ s-\dfrac{d}{336\ \dfrac{m}{s}},$$

here, $d$ is the distance from the top of the well to the surface of water.

Then, we can find the distance from the top of the well to the surface of water from the kinematic equation:

$$d=\dfrac{1}{2}gt^2,$$ $$d=\dfrac{1}{2}\cdot 10\ \dfrac{m}{s^2}\cdot(2.4\ s-\dfrac{d}{336\ \dfrac{m}{s}})^2,$$ $$5d^2-120960d+3251404.8=0.$$

This quadratic equation has two roots: $d_1=26.91\ m$ and $d_2=24165\ m$. Since the distance from the top of the well to the surface of water can't be such a big, we accept the first root and the distance will be $d=26.91\ m$.

b) if the travel time for the sound is ignored, we get:

$$d=\dfrac{1}{2}\cdot 10\ \dfrac{m}{s^2}\cdot(2.4\ s)^2=28.8\ m.$$

Finally, we can calculate the percent error:

$$\%error=\dfrac{28.8\ m-26.91\ m}{26.91\ m}\cdot 100\%=7\%.$$

Answer:

a) $d=26.91\ m$.

b) $\%error=7\%.$