Question

A man walks at the rate of $3\text{km/hr}$. Rain appears to him falling in vertical direction at the rate of $3\sqrt{3}\text{km/hr}$. Which of the following options is correct regarding the actual velocity of rain?

• A. $6\text{km/hr}$, inclined at an angle of ${30}^{\circ }$ to the vertical towards the man's motion.
• B. $3\text{km/hr}$, inclined at an angle of ${30}^{\circ }$ to the vertical towards the man's motion.
• C. $6\text{km/hr}$, inclined at an angle of ${45}^{\circ }$ to the vertical towards the man's motion.
• D. $6\text{km/hr}$, inclined at an angle of ${60}^{\circ }$ to the vertical towards the man's motion.

Answer 1

The correct option is A $6\text{km/hr}$, inclined at an angle of ${30}^{\circ }$ to the vertical towards the man's motion.

Let $\overrightarrow{v_r}$ be the actual velocity of rain.
Speed of rain w.r.t man, $v_{r/m}=3\sqrt{3}\text{km/hr}$
Speed of man w.r.t ground, $v_m=3\text{km/hr}$

$\overrightarrow{v_{r/m}}=\overrightarrow{v_r}-\overrightarrow{v_m}$
$\overrightarrow{v_r}=\overrightarrow{v_{r/m}}+\overrightarrow{v_m}$
$|\overrightarrow{v_r}|=\sqrt{v_{r/m}^2+v_m^2}$
$=\sqrt{(3{)}^2+(3\sqrt{3}{)}^2}=6\text{km/hr}$
$\tan \theta =\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^{\circ }$