Question

A man is coming down an incline of angle ${30}^{\circ }$. When he walks with speed $\sqrt{3}$m/s he has to keep his umbrella vertical w.r.t. ground to protect himself from rain. The actual speed of rain is 5 m/s. At what angle with vertical should he keep his umbrella when he is at rest so that he does not get drenched?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

${\overrightarrow{v}}_{man}=2\sqrt{3}cos{30}^{\circ }\hat{i}-2\sqrt{3}sin{30}^{\circ }\hat{j}$

${\overrightarrow{v}}_m=3\hat{i}-\sqrt{3}\hat{j}$

At this moment the rain appears to him as vertical.Let the magnitude of $|{\overrightarrow{v}}_{\frac{r}{m}}|=v$

$\Rightarrow {\overrightarrow{v}}_{\frac{r}{m}}=v\hat{j}$

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

$\Rightarrow {\overrightarrow{v}}_r={\overrightarrow{v}}_{\frac{r}{m}}+{\overrightarrow{v}}_m$

Draw vector diagram

$\Rightarrow |{\overrightarrow{v}}_r|=5$

Given now from it we know

${\overrightarrow{v}}_r=5sin\theta \hat{i}-5cos\theta ‘\hat{j}$

${\overrightarrow{v}}_r={\overrightarrow{v}}_{\frac{r}{m}}+{\overrightarrow{v}}_m$

$\Rightarrow 5sin\theta \hat{i}=-v\hat{j}+3\hat{i}-3\hat{j}$

$\Rightarrow (5sin\theta =3)$ -------------(1)

$-5cos\theta =-(v+3)$

From equation (1) we know $sin\theta =\frac{3}{5}$

$\Rightarrow \theta ={37}^{\circ }$