Question

Rain is falling vertically with a speed of $12m{s}^{-1}$.A cyclist is moving east to west with a speed of $12\sqrt{3}m{s}^{-1}$.In order to protect himself from rain at what angle he should hold his umbrella w.r.t vertical.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

${\overrightarrow{v}}_{rain}=-12\hat{j},{\overrightarrow{v}}_{cyclist}=-12\sqrt{3}\hat{i}$

$\therefore {\overrightarrow{v}}_{rc}={\overrightarrow{v}}_r-{\overrightarrow{v}}_c$

= $-12\hat{j}-(-12\sqrt{3}\hat{i})$

= $-12\hat{j}+12\sqrt{3}\hat{i}=12\sqrt{3}\hat{i}-12\hat{j}$

$\therefore tan\theta =\frac{12}{12\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\therefore \theta ={30}^{\circ }$

$\therefore$ Since the $v_{rain}$ w.r.t cyclist make an angle of ${30}^{\circ }$ with the x axis, he must hold the umbrella at an angle of ${60}^{\circ }$ to the vertical.