question_answer

A man is walking due east at the rate of 2 km/h. The rain appears to him to come down vertically at the rate of 2 km/h. The actual velocity and direction of rainfall with the vertical respectively are

A)  \[2\sqrt{2}\,km/h,{{45}^{0}}\]                 

B)  \[\frac{1}{\sqrt{2}}\,km/h,{{30}^{0}}\]

C)  \[2\,km/h,{{0}^{0}}\]                   

D)  \[1\,km/h,{{90}^{0}}\]

Correct Answer: A

Solution :

Velocity of man, \[{{v}_{m}}=2\,km/h\] Velocity of rain, \[{{v}_{r}}=2\,km/h\] Then actual velocity \[{{v}_{rm}}=\sqrt{v_{m}^{2}+v_{r}^{2}}\] \[=\sqrt{{{(2)}^{2}}+{{(2)}^{2}}}\] \[=\sqrt{4+4}=2\sqrt{2}\,km/h\] The direction of rainfall with the vertical \[\sin \theta =\frac{{{v}_{r}}}{{{v}_{rm}}}\]          \[=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\] \[\theta ={{\sin }^{-1}}\left( \frac{1}{\sqrt{2}} \right)\] \[\theta ={{45}^{o}}\]