It is raining vertically downwards with the velocity of $3 k m h^{- 1}$ . A man walks in the rain with a velocity of $4 k m h^{- 1}$ . The rain drops will be falling on the man with a relative velocity of

1 k m h^{- 1}

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3 k m h^{- 1}

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4 k m h^{- 1}

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5 k m h^{- 1}

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Solution

The correct option is A $5 k m h^{- 1}$
Given:
Vertical component $= 3 k m h^{- 1}$
Horizontal component (velocity of man) $= 4 k m h^{- 1}$
Resultant $= \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{9 + 16} = 5 k m h^{- 1}$
Hence relative velocity of rain on the man is $5 k m h^{- 1}$
Hence option $D$ is correct.

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