Question

Rain appears to fall vertically downwards to a man walking at the rate of $3\text{km/h}$. When he increases his speed to $6\text{km/h}$, the rain appears coming to him at an angle of ${45}^{\circ }$ with the vertical. The speed of rain is

• A. $3\sqrt{3}\text{km/h}$
• B. $3\sqrt{2}\text{km/h}$
• C. $3\sqrt{5}\text{km/h}$
• D. $5\sqrt{3}\text{km/h}$

Answer 1

The correct option is B $3\sqrt{2}\text{km/h}$

Let the velocity of rain of rain with respect to ground be, ${\overrightarrow{V}}_r=(x\hat{i}-y\hat{j})\text{km/h}$
Case I: Velocity of man w.r.t ground is,${\overrightarrow{V}}_m=3\hat{i}\text{km/h}$
Velocity of rain w.r.t man is given by,
${\overrightarrow{V}}_{rm}={\overrightarrow{V}}_r-{\overrightarrow{V}}_m$
${\overrightarrow{V}}_{rm}=(x-3)\hat{i}-y\hat{j}...\left(1\right)$


As rain appears to fall vertically downwards,
$\therefore \theta ={90}^{\circ }$ with $X-$axis
$\Rightarrow \tan \theta =\infty$
From Eq.$\left(1\right)$
$\left|\frac{y}{x-3}\right|=\infty$
i.e $x-3=0$
$\therefore x=3...\left(2\right)$

Case II: When velocity of man becomes, ${\overrightarrow{V}}_m=6\hat{i}\text{km/h}$
$\Rightarrow {\overrightarrow{V}}_{rm}={\overrightarrow{V}}_r-{\overrightarrow{V}}_m$
$=(x\hat{i}-y\hat{j})-6\hat{i}=(x-6)\hat{i}-y\hat{j}$
Substituting value of $x$ gives,
${\overrightarrow{V}}_{rm}=-3\hat{i}-y\hat{j}$
As rain appears to make ${45}^{\circ }$ with vertical,


$\tan {45}^{\circ }=\frac{|-y|}{|-3|}=\frac{y}{3}$
$\Rightarrow 1=\frac{y}{3}$
$\therefore y=3$
Therefore, velocity of rain w.r.t ground is
${\overrightarrow{V}}_r=3\hat{i}-3\hat{j}\text{km/h}$
Now, speed of rain or its magnitude is,
$V_r=\sqrt{(3{)}^2+(-3{)}^2}$
$\therefore V_r=3\sqrt{2}\text{km/h}$
Speed of the rain with respect to ground is $3\sqrt{2}\text{km/h}$.