Question

A standing man observes rain falling with the velocity of $20m{s}^{-1}$ at an angle of ${30}^o$ with the vertical.
a. Find the velocity with which the man should move so that rain appears to fall vertically to him.
b. Now if he further increases his speed, rain again appears to fall at ${30}^o$ with the vertical. Find his new velocity.

Answer 1

a. Velocity of man, ${\overrightarrow{v}}_m=-v\hat{i}\left(let\right)$

Velocity of rain, ${\overrightarrow{v}}_r=(-10\hat{i}-10\sqrt{3}\hat{j})m{s}^{-1}$

Velocity of rain w.r.t to man, = ${\overrightarrow{v}}_{r.m}={\overrightarrow{v}}_r-{\overrightarrow{v}}_m$

$=(-10\hat{i}-10\sqrt{3}\hat{j})-(-v\hat{i})m{s}^{-1}$

${\overrightarrow{v}}_{r.m}=-(10-v)\hat{i}-10\sqrt{3}\hat{j}m{s}^{-1}$

For vertical fall, the horizontal component must be zero.

$\Rightarrow -(10-v)=0$ or $v=10m{s}^{-1}$

b. Velocity of rain will not change hence,

${\overrightarrow{v}}_r=(-10\hat{i}-10\sqrt{3}\hat{j})m{s}^{-1}$

Let the velocity of man in this case be = $\overrightarrow{v_m}=-v_1\hat{i}$

Velocity of rain w.r.t. to man

${\overrightarrow{v}}_{r.m}={\overrightarrow{v}}_r-{\overrightarrow{v}}_m$

$=(-10\hat{i}-10\sqrt{3}\hat{j})-(-v_1\hat{i})m{s}^{-1}$

$=-(10-v_1)\hat{i}-10\sqrt{3}\hat{j}m{s}^{-1}$

Angle with the vertical = ${30}^o$

$\Rightarrow tan{30}^o=\frac{10-v_x}{-10\sqrt{3}}\Rightarrow v_x=20m{s}^{-1}$