To a man walking at the rate of 2 kmph, rain appears to fall vertically. When he doubles his speed, it appears to fall at 30° to the vertical. The actual velocity of the rain is

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8 kmph at 60° to the horizontal

4 kmph at 60° to the horizontal

4 kmph at 60° to the vertical

8 kmph at 60° to the vertical

answer is B.

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Detailed Solution

$\begin{array}{l} \tan 30^{\circ} = \frac{V}{y} \\ \frac{1}{\sqrt{3}} = \frac{2}{y} \\ \Rightarrow y = 2 \sqrt{3} \\ V_{rain} = \sqrt{x^{2} + y^{2}} \\ = \sqrt{4 \times 3 + 4} \\ = 4 kmph \\ V_{rain} = 4 kmph \\ \tan ϕ = \frac{2 \sqrt{3}}{2} \\ = \sqrt{3} \\ ϕ = 60^{\circ} \end{array}$

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