Question

A submarine, 608m below sea level, sends a sonar signal to detect an aircraft directly above it. If it receives the signal 10 sec. later, how far is the aircraft from the submarine? (Speed of sound in air 340 m/s, speed of sound in seawater= 1520 m/s)

Answer 1

Use the relation between velocity, distance, and time to determine the distance of the aircraft from the surface of the water based on the time the sound waves take to reach the aircraft. The received signal by the submarine has travelled the distance between the aircraft and the submarine twice.
Formula used:
$v = \dfrac{d}{t}$ where $v$ is the velocity of the sound wave, $d$ is the distance it travels in time $t$

Complete step by step answer:
Since we’ve given that the signal sent by the submarine is received in 10 secs, the time taken by the sound signal to reach the airplane from the submarine will be $t = 10/2 = 5\,\sec$
Let the distance between the submarine and the airplane be $x$ . Then using the relation $v = \dfrac{d}{t}$ , we can write the total time taken by the sound wave to travel from the submarine to the airplane as the sum of the times spent by the sound wave in the air and water. So,
$t = \dfrac{{{d_{water}}}}{{{v_{water}}}} + \dfrac{{{d_{air}}}}{{{v_{air}}}}$ where ${d_{water}} = 608\,m$ and ${d_{air}}$ is the height of the airplane above the seas level.
So, on substituting ${v_{air}} = 340\,m/s$ and ${v_{water}} = 1520\,m/s$ , we get
$5 = \dfrac{{608}}{{1520}} + \dfrac{{{d_{air}}}}{{340}}$
On rearranging the terms, we get:
$\dfrac{{{d_{air}}}}{{340}} = 5 - \dfrac{{608}}{{1520}}$
$\dfrac{{{d_{air}}}}{{340}} = 4.6$
Multiplying both sides by 340, we get
${d_{air}} = 1546\,m$
Then, the distance between the submarine and the airplane is:
$x = {d_{water}} + {d_{air}}$
$\therefore x = 1546 + 608 = 2172\,m$
Hence the airplane is situated 2172 meters above the submarine.

Note:
SONAR systems calculate the time taken by a sound wave to get reflected from the airplane and come back. However when we want to calculate the distance, we must take half of the measured time since the sound waves only take half of the total time in reaching the airplane.