Question

A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. If the speed of sound is 340 m/s, how high is the cliff?

Short answer

The height of the cliff is 51.76 m.

Step-by-step solution

Step 1. Meaning of free fall

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.

Step 2. Given information

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.

Step 3. Determination of time taken by the sound of the rock striking the ocean surface to reach the cliff

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.

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

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

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

Step 4. Determination of the time the rock fell

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.

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

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.

Step 5. Determination of the height of the cliff

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.