91Ó°ÊÓ

Here the rock is thrown from up; hence, it is a case of free falling body.

Let’s find out the known and the unknown value.

Initial velocity of the rock is zero as it starts its motion from stationary point.

The acceleration of the rock is 9.81 m/s².

The time taken by the rock to reach the surface of the water is 2 s.

Hence the known values are

\[\begin{array}{*{20}{l}}{U = 0}\\{a = {\rm{ }}9.81{\rm{ }}m/{s^2}}\\{t = 2{\rm{ }}s}\\{d = ?}\end{array}\]

Hence the distance can be calculated by

\(\begin{array}{c}d = ut + \frac{1}{2}g{t^2}\\d = \left( 0 \right)\left( 2 \right) + \frac{1}{2}\left( {9.81} \right){\left( 2 \right)^2}\\d = \,19.62\,\,m\,\end{array}\)

Hence the distance of the water from the surface is \[19.62 m\]

If we take in account the total time for the sound to travel up the well hence the equation to find out the time of the rock fall is

The distance of the rock touching the ground will be equal to the distance travelled by the sound wave.

Sound do not experience any acceleration as it is massless.

Hence by putting the values in the equation we get the below equation.

\(\begin{array}{l}{D_{rock}} = {D_{sound\,wave}}\\{u_r}{t_r} + \frac{1}{2}{g_r}{t_r}^2 = {u_s}{t_s} + \frac{1}{2}{g_s}{t_s}^2\\\left( 0 \right){t_r} + \frac{1}{2}\left( {9.81} \right){t_r}^2 = \left( {332.0} \right){t_s} + \frac{1}{2}\left( 0 \right){t_s}^2\\4.905{t_r}^2 = 332.0{t_s}\end{array}\)

If we take in account the total time for the sound to travel up the well hence the equation to find out the time of the rock fall is