Question

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 $0.5ms^{- 1},$ he finds that rain falls vertically. The speed of rain with respect to stationary man is

• A. $0.5ms^{- 1}$
• B. $1 \, m \, s^{- 1}$
• C. $\frac{\sqrt{3}}{2}ms^{- 1}$
• D. $\frac{1}{\sqrt{3}}ms^{- 1} \, $

Answer 1

Answer: (B)

For stationary person

For running person

Speed of rain with respect to a stationary man $= \nu$
$\overset{ \rightarrow }{v}=vsin30^{o}\hat{i}-vcos30^{o}\hat{j}$
The velocity of rain with respect to man,
$\overset{ \rightarrow }{v}_{R M}=\overset{ \rightarrow }{v}_{R}-\overset{ \rightarrow }{v}_{M}$
$=\left(\right. v sin \left(30\right)^{o} \hat{i} - v cos \left(30\right)^{o} \hat{j} \left.\right)-\left(\right. 0.5 \hat{i} \left.\right)$
Since $\overset{ \rightarrow }{V}_{R M}$ is vertically downward, its $x$ -component will become zero.
$\therefore vsin30^{o}-0.5=0$
$\Rightarrow v=\frac{0.5}{sin 30^{o}}=1ms^{- 1}$