Question

A man moving at speed of $5\text{m/s}$ observes rain falling vertically with velocity $5\sqrt{3}m/s$. Find the speed and direction of the rain w.r.t ground.

• A. $10\text{m/s}$, ${60}^{\circ }$ with vertical
• B. $10\text{m/s}$, ${30}^{\circ }$ with vertical
• C. $5\text{m/s}$, ${60}^{\circ }$ with vertical
• D. $5\text{m/s}$, ${30}^{\circ }$ with vertical

Answer 1

The correct option is B $10\text{m/s}$, ${30}^{\circ }$ with vertical

Given,
${\overrightarrow{v}}_{r,m}=-5\sqrt{3}\hat{j}\text{m/s}$
$\overrightarrow{v_m}=5\hat{i}\text{m/s}$
${\overrightarrow{v}}_{r,m}={\overrightarrow{v}}_r-{\overrightarrow{v}}_m$
${\overrightarrow{v}}_r={\overrightarrow{v}}_{r,m}+{\overrightarrow{v}}_m\Rightarrow {\overrightarrow{v}}_r=(-5\sqrt{3})\hat{j}+5\hat{i}\text{m/s}$
$|{\overrightarrow{v}}_r|=\sqrt{(-5\sqrt{3}{)}^2+5^2}$
$=\sqrt{75+25}=\sqrt{100}\text{m/s}$
$|{\overrightarrow{v}}_r|=10\text{m/s}$ (speed of main w.r.t. ground)
Inclination of the rain with vertical,
$\tan \theta =\frac{5}{5\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\theta ={\tan }^{-1}\frac{1}{\sqrt{3}}={30}^{\circ }$
$\theta ={30}^{\circ }$ with vertical