To a stationary man, rain appears to be falling at his back at an angle ${30}^o$ with the vertical. As he starts moving forward with a speed of $0.5 \mathrm{m} {\mathrm{s}}^{-1},$ he finds that rain falls vertically. The speed of rain with respect to stationary man is

• A

  $0.5 \mathrm{m} {\mathrm{s}}^{-1}$
• Correct answer
  B

  $1 \mathrm{m} {\mathrm{s}}^{-1}$
• C

  $\frac{\sqrt{3}}{\mathrm{2 }} \mathrm{m} {\mathrm{s}}^{-1}$
• D

  $\frac{1}{\sqrt{3}} \mathrm{m} {\mathrm{s}}^{-1}\mathrm{ }$

Solution

For stationary person

For running person

Speed of rain with respect to a stationary man $=\nu$
$\overrightarrow{v}=v \sin {30}^o \hat{\mathrm{i}}-v \cos {30}^o \hat{\mathrm{j}}$
The velocity of rain with respect to man,
${\overrightarrow{v}}_{RM}={\overrightarrow{v}}_R-{\overrightarrow{v}}_M$
$=(v \sin {30}^o \hat{\mathrm{i}}-v \cos {30}^o \hat{\mathrm{j}})-(0.5 \hat{\mathrm{i}})$
Since ${\overrightarrow{V}}_{RM}$ is vertically downward, its $x$-component will become zero.
$\therefore v \sin {30}^o-0.5=0$
$\Rightarrow v=\frac{0.5}{\sin {30}^o}=1 \mathrm{m} {\mathrm{s}}^{-1}$

Hence, the correct answer is option: 'B'.