Question

A person standing on open ground hears the sound of a jet airplane, coming from the north at an angle ${30}^{\circ }$ with the horizontal. But he finds the airplane 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. $\frac{2}{\sqrt{3}}v$
• D. $\frac{v}{2}$

Answer 1

The correct option is A $\frac{\sqrt{3}}{2}v$

Let the height of airplane be $d$ from the ground.


As sound produced by the airplane at point $A$ reaches the man at point $O$, the airplane moves to point $B$ and will be vertically above the man.
Time taken by sound to reach the man
$t=\frac{d}{v_{\perp }}$
$t=\frac{d}{v\sin {30}^{\circ }}...\left(1\right)$
In the same time, the airplane will go from point $A$ to $B$, hence speed of the airplane is,
$v_0=\frac{AB}{t}$


Substituting from the triangle shown and Eq.$\left(1\right)$,
$v_0=\frac{\left(\frac{d}{\tan {30}^{\circ }}\right)}{\left(\frac{d}{v\sin {30}^{\circ }}\right)}$
$\Rightarrow v_0=\frac{\sqrt{3}v}{2}$
The speed of the plane is $\frac{\sqrt{3}v}{2}$.