Question

A person standing on a road has to hold his umbrella at 60${}^{\circ }$ with the vertical to keep away rain. He throws the umbrella and starts running in horizontal direction at a speed of $20\text{m/s}$. He finds that the rain drops are hitting his head vertically. Find the speed of rain drops with respect to ground.

• A. $\frac{40}{\sqrt{3}}\text{m/s}$
• B. $11.5\text{m/s}$
• C. $\frac{20}{\sqrt{3}}\text{m/s}$
• D. $17\text{m/s}$

Answer 1

The correct option is A $\frac{40}{\sqrt{3}}\text{m/s}$

While at rest person is holding the umbrella at 60${}^{\circ }$ to avoid rain.
$\therefore {\overrightarrow{v}}_{rg}=$ velocity of rain w.r.t ground, i.e at ${60}^{\circ }$ from vertical.
After man runs horizontally the rain appears vertically downwards to him.
${\overrightarrow{v}}_{rm}=$ velocity of rain w.r.t man
${\overrightarrow{v}}_{mg}=20\hat{i}\text{m/s}$

From relative velocity definition:
${\overrightarrow{v}}_{rg}={\overrightarrow{v}}_{rm}+{\overrightarrow{v}}_{mg}$

From the vector triangle shown in figure:
$\sin \theta =\frac{|{\overrightarrow{v}}_{mg}|}{|{\overrightarrow{v}}_{rg}|}$

$\therefore \sin {60}^{\circ }=\frac{20}{|{\overrightarrow{v}}_{rg}|}$
$|{\overrightarrow{v}}_{rg}|=\frac{40}{\sqrt{3}}$

$\therefore |{\overrightarrow{v}}_{rg}|=\frac{40}{\sqrt{3}}\text{m/s}$
$|{\overrightarrow{v}}_{rg}|$ is the speed of rain w.r.t ground.