Question

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

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

Answer 1

Answer: (B)

$3\sqrt{2}km/h$

Let $\hat{i}and\hat{j}$ be the unit vectors in horizontal and vertical directions respectively.
Let velocity of rain
$\overrightarrow{v}=a\hat{i}+b\hat{j}$ ...(i)
Then speed of rain will be $\left|{\overrightarrow{v}}_r\right|=\sqrt{a^2+b^2}$ ...(ii)
In the first case ${\overrightarrow{v}}_m$ = velocity of man = $3\hat{i}$
$\therefore {\overrightarrow{v}}_{nn}={\overrightarrow{v}}_r-{\overrightarrow{v}}_m=(a-3)\widehat{i+b\hat{j}}$
It seems to be in vertical direction. Hence, a-3=0 or a=3
In the second case ${\overrightarrow{v}}_m=6\hat{i}$
$\therefore {\overrightarrow{v}}_{nn}=(a-6)\hat{i}+b\hat{j}=3\hat{i}+b\hat{j}$
This seems to be at ${45}^{\circ }$ with vertical.
Hence, $\left|b\right|=3$
Therefore, from Eq. (ii) speed of rain is
$\left|{\overrightarrow{v}}_r\right|=\sqrt{{\left(3\right)}^2+{\left(3\right)}^2}=3\sqrt{2}km/h$