Question

A person standing on an open ground hears the sound of a jet aeroplane coming from north at an angle ${60}^{\circ }$ with ground level. But he finds the aeroplane right vertically above his position. If $v$ is the speed of sound, then speed of the plane is

• A. $\frac{\sqrt{3}}{2}v$
• B. $v$
• C. $\frac{2v}{\sqrt{3}}$
• D. $\frac{v}{2}$

Answer 1

The correct option is D $\frac{v}{2}$

Let $v_P$ and $v$ be the speed of the plane and speed of the sound.
Let $P_1$ be the position of plane at $t=0$,
when sound waves started towards person $A$, $P_2$ is the position of plane observed at time instant $t$ as shown in the figure below.

In triangle $P_1{P}_{2}A.$
$P_1{P}_2$ = speed of plane $\times$ time $=v_P\times t$
$P_{1}A$ = speed of sound $\times$ time $=v\times t$
Now, from $\Delta P_1{P}_{2}A$,
$\cos {60}^{\circ }=\frac{P_1{P}_2}{P_{1}A}=\frac{v_P\times t}{v\times t}$
$\frac{1}{2}=\frac{v_P}{v}$
$v_P=\frac{v}{2}$