Question

A man running along a straight road with uniform velocity $\overrightarrow{u}=u\hat{i}$ feels that the rain is falling vertically down along $-\hat{j}$. If he doubles his speed,he finds that the rain is coming at an angle $\theta$ with the vertical. The velocity of the rain with respect to the ground is :

• A. $ui-u\tan \theta \hat{j}$
• B. $ui-\frac{u}{tan\theta }\hat{j}$
• C. $u\tan \theta -u\overrightarrow{j}$
• D. $\frac{u}{\tan \theta }\overrightarrow{i}-u\overrightarrow{j}$

Answer 1

The correct option is B $ui-\frac{u}{tan\theta }\hat{j}$

Given,

Velocity of man $=\overrightarrow{u}=u\hat{i}$

Let the velocity of rain $=\overrightarrow{v}=x\hat{i}+y\hat{j}$

Now, in first case velocity of rain with respect

to man $=\overrightarrow{V_R}=\overrightarrow{v}-\overrightarrow{u}$

$=(x-u)\hat{i}+y\hat{j}$

Given that It $x-$ component is zero as

rain is falling vertically down, and $y$ component

is along $-ve$ direction

So, $x-u=0$

$\Rightarrow x=u$

$\therefore$ velocity of rain $=u\hat{i}+y\hat{j}$

In second case when be will doobles his speed velocity of rain with respect to man $=\overrightarrow{V_R}$

$\Rightarrow \overrightarrow{V_R}=\overrightarrow{v}-\overrightarrow{u}$

$=u\hat{i}+y\hat{j}-2u\hat{i}$

$=-u\hat{i}+y\hat{j}$

Now this case rain is coming at an angle $\theta$ to the vertical

$\therefore \tan \theta =\frac{-u}{y}$

$\Rightarrow y=\frac{-u}{\tan \theta }$

$\therefore$ velocity of rain with respect to ground

$=\overrightarrow{V}=x\hat{i}+y\hat{j}$

$=u\hat{i}-\frac{y}{\tan \theta }\hat{j}$

$\therefore$ Option $B$ is correct