Answer to Question #94877 in Mechanics | Relativity for Erykah

Question

A torpedo fired from a submerged submarine is propelled through the water with a speed of 20 m/s and explodes on impact with a target 2000m away. If the sound of the impact is heard 101.4 s after the torpedo was fired, what is the speed of sound in water?

Answer 1

Let’s denote $u$ – speed of sound, $v$ – speed of torpedo and $l$ – initial distance between submarine and target. The total amount of time $t$ is the time taken by the torpedo to hit the target plus the time taken by sound to reach the submarine. Assuming uniform movement of torpedo and neglecting the movement of target we get (the distance traveled by the torpedo and sound are equal)

$$t = \frac{l}{v} + \frac{l}{u}$$

Now let’s solve this equation (leading left part to common denominator and selecting the term contain speed of sound)

$$\frac{l}{u} = \frac{{vt - l}}{v}$$

Dividing by $l$ and flipping the fractions we finally get

$$u = \frac{{vl}}{{vt - l}}$$

Now let’s calculate (units are shown in square brackets)

$$u = \frac{{20[\frac{m}{s}] \cdot 2000[m]}}{{20[\frac{m}{s}] \cdot 101.4[s] - 2000[m]}} = 1428.57[\frac{m}{s}]$$

This is the answer.