Question

A person standing on an open ground hears the sound of a jet aeroplane, coming from the north at an angle ${60}^0$ with ground level. But he finds the aeroplane right vertically above his position. If $v$ is the speed of sound, the speed of the plane is

• A. $\left(\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)v$
• B. $\raisebox{1ex}{$2v$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.$
• C. $V$
• D. $\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Answer 1

The correct option is D

$\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Step 1. Given Data:

Angle $\theta ={60}^{°}$

Speed of sound $=$ $v$

Speed of plane =$v_p$

Let the person is standing on point $R$, he hears the sound of plane from point $Q$ at angle $\theta ={60}^{°}$and finds the plane vertically above at his position at point $P$.

Step 2. From the figure:

We know that,

(Distance $=$Speed $\times$Time)

Distance,$PQ=v_p\times t$

Distance, $QR=v\times t$

Step 3. To calculate the speed of plane $v_p$:

Since he hears the sound of plane from point $Q$ at angle $\theta ={60}^{°}$

$Cos{60}^0=\frac{PQ}{QR}$ $\left(\because Cos\theta =\frac{B}{H}\right)$

$$\begin{gathered} \Rightarrow \frac{1}{2}=\frac{(v_p\times t)}{(v\times t)} \\ \therefore v_p=\frac{v}{2} \end{gathered}$$

Hence, the correct option is (D).