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A man standing on an inclined plane observes that rain is falling vertically. When he starts moving down the inclined plane with velocity $V_{m} = 6 m / s$ , observes that rain hitting him horizontally. Then actual velocity of rain is

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$3 \sqrt{3} m / s$

3 m/s

4 m/s

$4 \sqrt{3} m / s$

answer is B.

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Detailed Solution

${\vec{V}}_{R} = {\vec{V}}_{m} + {\vec{V}}_{\frac{R}{m}}$

$V_{R} = V_{m} \sin 60^{o} = 6 \times \frac{\sqrt{3}}{2} m / s = 3 \sqrt{3} m / s$

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