Question

To a man running at the rate of $4\text{km/h}$, the rain appears to fall vertically. When he increases his speed to $8\text{km/h}$, rain appears coming to man at an angle of ${45}^{\circ }$ with vertical. Find the magnitude of velocity of rain with respect to ground.

• A. $3\sqrt{2}\text{km/h}$
• B. $4\sqrt{2}\text{km/h}$
• C. $2\sqrt{2}\text{km/h}$
• D. $5\sqrt{2}\text{km/h}$

Answer 1

The correct option is B $4\sqrt{2}\text{km/h}$

Let the velocity of the rain w.r.t ground be ${\overrightarrow{v}}_{rg}=a\hat{i}+b\hat{j}\text{km/h}$
where $\hat{i}$ and $\hat{j}$ are the unit vector in horizontal and vertical direction respectively.

Magnitude of velocity of rain w.r.t ground $|{\overrightarrow{v}}_{rg}|=\sqrt{a^2+b^2}\text{km/h}...\left(1\right)$

In first case,
Velocity of man w.r.t ground ${\overrightarrow{v}}_{mg}=4\hat{i}\text{km/h}$
${\overrightarrow{v}}_{rm}={\overrightarrow{v}}_{rg}-{\overrightarrow{v}}_{mg}=(a-4)\hat{i}+b\hat{j}$
$\because$Rain appear to man to be in vertical direction, $x-$component of ${\overrightarrow{v}}_{mg}$ should be zero.
$\Rightarrow (a-4)=0\Rightarrow a=4$
In second case;
${\overrightarrow{v}}_{mg}=8\hat{i}\text{km/h}$
${\overrightarrow{v}}_{rm}=(a-8)\hat{i}+b\hat{j}$ (but a = 4)
${\overrightarrow{v}}_{rm}=-4\hat{i}+b\hat{j}$

In this case, rain appears to be at ${45}^o$ with vertical.
$\tan {45}^{\circ }=\frac{b}{-4}\Rightarrow b=-4$
$|b|=4$
Magnitude of velocity of rain w.r.t ground
$|{\overrightarrow{v}}_{mg}|=\sqrt{a^2+b^2}$
$\therefore |{\overrightarrow{v}}_{mg}|=\sqrt{4^2+4^2}=4\sqrt{2}\text{km/h}$