Question

A man moving with velocity $5\text{m/s}$ along a straight line observes rain falling vertically at the rate of $10\text{m/s}$. Find the speed of rain with respect to ground, and angle$\left(\theta \right)$ made from vertical by the rain as seen from ground.

• A. $5\sqrt{5}\text{m/s},\theta =\frac{{\tan }^{-1}4}{2}$
• B. $5\sqrt{5}\text{m/s},\theta ={\tan }^{-1}\frac{1}{2}$
• C. $10\text{m/s},\theta ={\tan }^{-1}2$
• D. $10\text{m/s},\theta ={\tan }^{-1}\frac{1}{2}$

Answer 1

The correct option is B $5\sqrt{5}\text{m/s},\theta ={\tan }^{-1}\frac{1}{2}$

Velocity of man with respect to ground
${\overrightarrow{v}}_{MG}=5\hat{i}\text{m/s}$

Velocity of rain with respect to man
${\overrightarrow{v}}_{RM}=-10\hat{j}\text{m/s}$

Velocity of rain with respect to ground is:

${\overrightarrow{v}}_{RG}={\overrightarrow{v}}_{RM}+{\overrightarrow{v}}_{MG}$

$\Rightarrow {\overrightarrow{v}}_{RG}=-10\hat{j}+5\hat{i}\text{m/s}$

$\therefore$Speed of rain with respect to ground:

$|{\overrightarrow{v}}_{RG}|=\sqrt{(-10{)}^2+5^2}=5\sqrt{5}\text{m/s}$

From the vector triangle, angle made by rain from vertical $\theta$ is given as:


$\tan \theta =\frac{|{\overrightarrow{v}}_{MG}|}{|{\overrightarrow{v}}_{RM}|}$

$\Rightarrow \tan \theta =\frac{5}{10}=\frac{1}{2}$

$\therefore \theta ={\tan }^{-1}\frac{1}{2}$