Question

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?

Short answer

The height of the cliff is $33.24\mathrm{m}$.

Step-by-step solution

Understanding the concept of distance travelled by system during free fall

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.

Given data

The time of striking heard is $T=2.7\mathrm{s}$.

Evaluating the value of travel time of sound to reach the top of cliff and time taken by stone to reach the water surface

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

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:

$\begin{array}{r}lT=t_1+t_2\\ t_1=T-t_2\\ t_1=T-\left(\frac{h}{v}\right)\end{array}$

Evaluating the height of a cliff by using the Newton’s equation of motion

The height of cliff is calculated below:

$\begin{array}{r}ch=u{t}_1+\frac{1}{2}g{\left(t_1\right)}^2\\ h=\left(0\right)t_1+\frac{1}{2}g{\left[T-\left(\frac{h}{v}\right)\right]}^2\\ h^2-2v\left(\frac{v}{g}+T\right)h+T^2{v}^2=0\end{array}$

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

Substitute the values in the above equation.

$\begin{array}{r}c{h}^2-2\left(343\mathrm{m}/\mathrm{s}\right)\left(\frac{343\mathrm{m}/\mathrm{s}}{9.81\mathrm{m}/{\mathrm{s}}^2}+2.7\mathrm{s}\right)h+{\left(2.7\mathrm{s}\right)}^2{\left(343\mathrm{m}/\mathrm{s}\right)}^2=0\\ h^2-25837.72h+857661.21=0\end{array}$

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.

$\begin{array}{r}ch=\frac{-\left(-25837.72\right)\pm \sqrt{{\left(-25837.72\right)}^2-4\left(1\right)\left(857661.21\right)}}{2\left(1\right)}\\ h=33.24\mathrm{m}\end{array}$

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 $33.24\mathrm{m}$.