Question

A man running on a horizontal road at $8m{s}^{-1}$ finds rain falling vertically. If he increases his speed to $12m{s}^{-1}$, he finds that drops make ${30}^{\circ }$ angle with the vertical. Find velocity of rain with respect to the road.

• A. $4\sqrt{7}m{s}^{-1}$
• B. $8\sqrt{2}m{s}^{-1}$
• C. $7\sqrt{3}m{s}^{-1}$
• D. $8.32m{s}^{-1}$

Answer 1

The correct option is D $8.32m{s}^{-1}$

let consider the velocity of rain,

${\overrightarrow{v}}_r=a\hat{i}+b\hat{j}$

${\overrightarrow{v}}_m=$ velocity of man

${\overrightarrow{v}}_{rm}=$ velocity of rain with respect to man

Condition (1): ${\overrightarrow{v}}_m=8\hat{i}$

${\overrightarrow{v}}_{rm}=0\hat{i}+b\hat{j}$

${\overrightarrow{v}}_r-{\overrightarrow{v}}_m=b\hat{j}$

$a\hat{i}+b\hat{j}-8\hat{i}=b\hat{j}$

$(a-8)\hat{i}+b\hat{j}=b\hat{j}$

$a-8=0$

$a=8$

Condition (2): ${\overrightarrow{v}}_m=12\hat{i}$

${\overrightarrow{v}}_{rm}={\overrightarrow{v}}_r-{\overrightarrow{v}}_m$

${\overrightarrow{v}}_{rm}=(a\hat{j}+b\hat{j})-12\hat{i}$

${\overrightarrow{v}}_{rm}=(a-12)\hat{i}+b\hat{j}$

Make ${30}^0$ angle to the vertical,

$\frac{b}{a-12}=tan{30}^0$

$\frac{b}{4-12}=\frac{1}{\sqrt{3}}$

$b=-\frac{4}{\sqrt{3}}$

So, velocity of rain with respect to the road is,

${\overrightarrow{v}}_r=8\hat{i}-\frac{4}{\sqrt{3}}\hat{j}$

It's magnitude, $|{\overrightarrow{v}}_r|=\sqrt{(8{)}^2+(-\frac{4}{\sqrt{3}}{)}^2}$

$v_r=8.32m/s$