Shown below is a submarine scouting an enemy ship in the ocean using a sonar device. Sonar devices send out…
Shown below is a submarine scouting an enemy ship in the ocean using a sonar device. Sonar devices send out a sound pulse from a transducer, and then precisely measure the time it takes for the sound pulses to be reflected back to the transducer.
A sonar wave sent by the submarine hits the ship and returns back in 2 seconds. The speed of a sonar wave underwater is 1500 m/s and the submarine is diving at a depth of 750 m below sea level.
Find the angle of elevation 'a' (without units) of the ship from the submarine (without symbol). Show your steps.
Solution
Step 1 — Total distance covered by the sonar wave.
The pulse travels from the submarine to the ship and back in 2 s at 1500 m/s.
Total distance $= 1500 \times 2 = 3000$ m.
Step 2 — Distance from the submarine to the ship.
That 3000 m is the round trip, so the one-way distance is
$BS = \dfrac{3000}{2} = 1500$ m.
Step 3 — Set up the right triangle.
Let $B$ be the submarine and $S$ the ship, and let $D$ be the point on the submarine's horizontal level directly below the ship. Then $\angle BDS = 90^\circ$, the vertical rise $SD = 750$ m (the submarine's depth below sea level), and the hypotenuse $BS = 1500$ m. The angle of elevation of the ship from the submarine is $a = \angle SBD$.
Step 4 — Find the angle.
$\sin a = \dfrac{SD}{BS} = \dfrac{750}{1500} = \dfrac{1}{2}$
$\sin a = \sin 30^\circ \Rightarrow a = 30^\circ$
Answer: 30
Asked in: NCERT Competency Based Questions (Class 10)
