Question

A man is coming down an inclined plane of angle ${30}^{\circ }$ with speed $2\sqrt{3}\text{m/s}$. He has to keep his umbrella vertical to protect himself from rain. If the actual magnitude of velocity of rain is $5\text{m/s}$, at what angle with vertical should he keep his umbrella when he is at rest?

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${37}^{\circ }$
• D. ${53}^{\circ }$

Answer 1

The correct option is C ${37}^{\circ }$

While coming down the inclined plane man has to keep his umbrella vertical $\Rightarrow$ rain is falling vertically w.r.t man
let Velocity of rain w.r.t. man$;V_{r,m}=x\hat{j}$ then

${\overrightarrow{V}}_{r,m}=x\hat{j}={\overrightarrow{V}}_r-{\overrightarrow{V}}_m$

Now velocity of man in vector form is
${\overrightarrow{V}}_m=V_m\cos {30}^{\circ }\hat{i}-V_m\sin 30\hat{j}=2\sqrt{3}\cos {30}^{\circ }\hat{i}-2\sqrt{3}\sin 30\hat{j}=3\hat{i}-\sqrt{3}\hat{j}$

Now, from above
${\overrightarrow{V}}_{r,m}={\overrightarrow{V}}_r-{\overrightarrow{V}}_m\Rightarrow x\hat{j}={\overrightarrow{V}}_r-(3\hat{i}-\sqrt{3}\hat{j}){\overrightarrow{V}}_r=3\hat{i}+(x-\sqrt{3})\hat{j}...\left(i\right)$

Given: $|{\overrightarrow{V}}_r|=5\text{m/s}$
So.
$\Rightarrow 5=\sqrt{3^2+(x-\sqrt{3}{)}^2}\Rightarrow 16=(x-\sqrt{3}{)}^2$

$\Rightarrow x=(4+\sqrt{3})\text{or}x=(-4+\sqrt{3})$
We will take $-ve$ value of $x$ as rain is falling vertically w.r.t man.
$\therefore V_{r,m}=(-4+\sqrt{3})\hat{j}...\left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$
$V_r=3\hat{i}+(x-\sqrt{3})\hat{j}$
${\overrightarrow{V}}_r=3\hat{i}-4\hat{j}$

As rain is falling with angle ${37}^{\circ }$ from vertical, man should hold umbrella at an angle ${37}^{\circ }$ with vertical while at rest.