91Ó°ÊÓ

When an object is allowed to fall freely under the effect of gravity, the object's motion is known as free fall, and the object moves with acceleration due to gravity. A stone dropped from some height is an example of free fall.

The speed of the sound is 340 m/s.

The time taken by the sound to reach the cliff after dropping of the rock is $t=3.4\mathrm{s}$.

Since a rock is dropped, its initial velocity is $u=0\mathrm{m}·{\mathrm{s}}^{-1}$.

The acceleration of the rock falling downward is equal to acceleration due to gravity, i.e.,$a=g=9.8\mathrm{m}·{\mathrm{s}}^{-2}$

Assume that the height of the cliff is s.

If t is the time taken by the rock to reach the ocean surface and $t\text{'}$is the time taken by the sound of rock striking the ocean surface to reach the cliff from the ocean surface, then.

role="math" localid="1643085151899" $t+t\text{'}=3.4\mathrm{s}...\left(\mathrm{i}\right)$

Since distance traveled is a product of speed and time, and the sound traveled with a speed of 340 m/s in time $t\text{'}$ to reach the cliff of height s, therefore.

$\begin{array}{l}

s = 340t'\\

t' = \frac{s}{{340}}

\end{array}$ ... (ii)

Use the third equation of motion for the downward motion of the rock.

$\begin{array}{l}s=ut+\frac{1}{2}a{t}^2\\ s=\left(0\right)t+\frac{1}{2}\left(9.8\right)t^2\\ s=4.9t^2…\left(iii\right)\end{array}$

On substituting the value of sfrom equation (3) in equation (2), you get.

$t\text{'}=\frac{4.9t^2}{340}...\left(iv\right)$

Substitute the value of $t\text{'}$from equation (4) in equation (1) to get the value of t.

$\begin{array}{c}t+\frac{4.9t^2}{340}=3.4\\ 4.9t^2+340t-1156=0\end{array}$

On solving the above quadratic equation for t, you will get.

$t=\frac{-340Â\pm \sqrt{{\left(340\right)}^2-4\left(4.9\right)\left(-1156\right)}}{2×4.9}$

The values of tare obtained as -72.64 s and 3.25 s. However, time cannot be negative. Thus, the appropriate value of t is 3.25 s.

So, the time taken by the rock to reach the ocean surface is 3.25 s.

On substituting 3.25 s for t in equation (3), you will get the value of s.

$\begin{array}{c}s=4.9{\left(3.25\right)}^2\\ =51.76\mathrm{s}\end{array}$

Thus, the height of the cliff is 51.76 s.