Question

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

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

Answer 1

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

Le the velocity of rain w.r.t ground, ${\overrightarrow{V}}_r=x\hat{i}+y\hat{j}$

Let
Case (i): When the man is walking at the rate of $3km/h$.
Relative velocity of rain w.r.t man,
${\overrightarrow{V}}_{rm}={\overrightarrow{V}}_r-{\overrightarrow{V}}_m=-(x\hat{i}+y\hat{j})+3\hat{i}=(3-x)\hat{i}-y\hat{j}$
Since rain appears to fall vertically,
$x-$ component $=0$
$\Rightarrow x=3$
Case (ii): When the man is walking at the rate of $6km/h$
${\overrightarrow{V}}_{rm}=(6-x)\hat{i}-y\hat{j}$
Now rain appears to fall at an angle of ${45}^{\circ }$ to the vertical.
$\Rightarrow \tan {45}^{\circ }=\frac{V_y}{V_x}=\frac{-y}{6-x}=\frac{y}{x-6}=\frac{y}{-3}$
(as $x=3$)
$\Rightarrow y=-3$
$\Rightarrow {\overrightarrow{V}}_r=3\hat{i}-3\hat{j}$
$|\overrightarrow{V}r|=\sqrt{3^2+(-3{)}^2}=3\sqrt{2}km/h$