A man is going due east with a velocity of 5 ms^-1. It is vertically raining downwards with a velocity of 4 ms^-1. At what angle should he hold the umbrella to the vertical so as to protect him self from the rain?

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$\tan^{- 1} (\frac{5}{4})$ in clock-wise direction

$\tan^{- 1} (\frac{4}{5})$ East of North

$\tan^{- 1} (\frac{4}{5})$ North of East

$\tan^{- 1} (\frac{5}{4})$ in anti-clockwise direction

answer is B.

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Detailed Solution

Relative to man, rain is coming from east at 5 km/h and from top at 4 km/h.

$Tan θ = \frac{v_{m}}{v_{r}}$ (clockwise) $θ = \tan^{- 1} (\frac{5}{4}) in clockwise direction$

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