Answer to Question #204515 in Physics for Nooria Haidari

Question

A student drops a rock into a water well. Exactly 2.472 seconds after she released the rock, she hears the sound of the splash caused by the rock hitting the water's surface. What is the depth of the water's surface from the top of the well? Assume that the speed of sound in the air to be 340 m/s.

Answer 1

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

$$t=2.472\ s-\dfrac{d}{340\ \dfrac{m}{s}},$$

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:

$$d=\dfrac{1}{2}gt^2,$$ $$d=\dfrac{1}{2}\cdot 10\ \dfrac{m}{s^2}\cdot(2.472\ s-\dfrac{d}{340\ \dfrac{m}{s}})^2,$$ $$5d^2-115600d+3523199=0.$$

This quadratic equation has two roots: $d_1=30.5\ m$ and $d_2=23089\ m$. 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 $d=30.5\ m$.