A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 2.7 s later. How high is the cliff?

During the free-fall motion, only a gravitational force is generating acceleration over the system. The distance travelled by the system during free fall is a function of the initial speed of the object, time of travel, and gravitational acceleration.

The time of striking heard is \(T = 2.7\;{\rm{s}}\).

The standard value for the speed of sound in air from Table-12.1 is \(v = 343\;{\rm{m/s}}\), and the standard value for gravitational acceleration is \(g = 9.81\;{\rm{m/}}{{\rm{s}}^2}\).

The travel time of sound to reach the top of cliff is calculated below:

\({t_2} = \frac{h}{v}\)

Here, h is the height of cliff.

The time taken by stone to reach the water surface is calculated below:

\(\begin{aligned}{l}T = {t_1} + {t_2}\\{t_1} = T - {t_2}\\{t_1} = T - \left( {\frac{h}{v}} \right)\end{aligned}\)

The height of cliff is calculated below:

\(\begin{aligned}{c}h = u{t_1} + \frac{1}{2}g{\left( {{t_1}} \right)^2}\\h = \left( 0 \right){t_1} + \frac{1}{2}g{\left[ {T - \left( {\frac{h}{v}} \right)} \right]^2}\\{h^2} - 2v\left( {\frac{v}{g} + T} \right)h + {T^2}{v^2} = 0\end{aligned}\)

Here, u is the initial speed of stone having a value of zero, g is the gravitational acceleration having a standard value of \(9.81\;{\rm{m/}}{{\rm{s}}^2}\).

Substitute the values in the above equation.

\(\begin{aligned}{c}{h^2} - 2\left( {343\;{\rm{m/s}}} \right)\left( {\frac{{343\;{\rm{m/s}}}}{{9.81\;{\rm{m/}}{{\rm{s}}^2}}} + 2.7\;{\rm{s}}} \right)h + {\left( {2.7\;{\rm{s}}} \right)^2}{\left( {343\;{\rm{m/s}}} \right)^2} = 0\\{h^2} - 25837.72h + 857661.21 = 0\end{aligned}\)

On solving the above quadratic equation,

\(h = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

Here, a, b and c are the equation constants for \(a{h^2} + bh + c = 0\).

Substitute the values in the above equation.

\(\begin{aligned}{c}h = \frac{{ - \left( { - 25837.72} \right) \pm \sqrt {{{\left( { - 25837.72} \right)}^2} - 4\left( 1 \right)\left( {857661.21} \right)} }}{{2\left( 1 \right)}}\\h = 33.24\;{\rm{m}}\end{aligned}\)

The largest root is impossible to consider because it will consume more time than 2.7 seconds to fall at that distance.

Hence, the height of cliff is \(33.24\;{\rm{m}}\).