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A man standing, observes rain falling with velocity of $20 m / s$ at an angle of $30^{°}$ with the vertical. Find out velocity of man so that rain again appears to fall at $30^{°}$ with the vertical.

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$30 m / s$

$40 m / s$

$20 m / s$

$10 m / s$

answer is A.

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Detailed Solution

$\vec{V_{R}} = 10 \hat{i} - 10 \sqrt{3} \hat{j};$

${\vec{V}}_{m} = V_{x} \hat{i};$

${\vec{V}}_{R M} = {\vec{V}}_{R} - {\vec{V}}_{m} = (10 - V_{x}) \hat{i} - 10 \sqrt{3} \hat{j}$

Angle with the vertical = $30^{0}$

$\tan 30^{°} = \frac{10 - V_{x}}{- 10 \sqrt{3}} \Rightarrow V_{x} = 20 m / s$

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