Question

A person walking at $3\text{kmph}$ finds rain falling vertically. When he walks in the opposite direction with the same speed, the rain appears to fall at ${45}^{\circ }$ to the vertical. What is the speed of rain ?

• A. $3\sqrt{5}\text{kmph}$
• B. $5\sqrt{3}\text{kmph}$
• C. $\frac{5}{\sqrt{3}}\text{kmph}$
• D. $\frac{3}{\sqrt{5}}\text{kmph}$

Answer 1

The correct option is A $3\sqrt{5}\text{kmph}$

Let velocity of rain with respect to earth be ${\overrightarrow{v}}_r=v_x\hat{i}+v_y\hat{j}$
Case (1):
Velocity of man ${\overrightarrow{v}}_{\text{m}}=3\hat{i}$
Relative velocity of rain w.r.t man ${\overrightarrow{v}}_{\text{rm}}=(v_x-3)\hat{i}+v_y\hat{j}$
Since, rain appears to fall vertically, $x$ component is zero
$\therefore v_x=3\text{kmph}$

Case (2):
Velocity of man ${\overrightarrow{v^{{}^{\prime }}}}_{\text{m}}=-3\hat{i}$
Relative velocity of rain w.r.t man
${\overrightarrow{v^{{}^{\prime }}}}_{\text{rm}}=(v_x+3)\hat{i}+v_y\hat{j}=6\hat{i}+v_y\hat{j}$
Since, rain appears to fall at ${45}^{\circ }$ to the vertical,
$\frac{v_y}{v_x}=\tan {45}^{\circ }=\frac{v_y}{6}\Rightarrow v_y=6\text{kmph}$
$\therefore$ Velocity of rain with respect to earth is
$\left|v_r\right|=\sqrt{v_x^2+v_y^2}=\sqrt{3^2+6^2}=3\sqrt{5}\text{kmph}$