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A man running at a speed of $5 \mathrm{~km} / \mathrm{h}$, find that the rain falls vertically. When he stops running, he finds that the rain is falling at an angle of $60^{\circ}$ with the horizontal. The velocity of rain with respect to running man is
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The correct answer is:
${5}{\sqrt{3}} \mathrm{~km} / \mathrm{h}$
Consider the situational diagram as shown below
$\begin{aligned} & \text { In } \triangle \mathrm{OAP}, \tan 30^{\circ}=\frac{v_m}{v_{r m}} \\ & \frac{1}{\sqrt{3}}=\frac{\text { Velocity of man }}{\text { Velocity of rain with respect to man }} \\ & \Rightarrow \quad v_{r m}=\frac{5}{1 / \sqrt{3}}=5 \sqrt{3} \mathrm{~km} / \mathrm{h}\end{aligned}$
$\begin{aligned} & \text { In } \triangle \mathrm{OAP}, \tan 30^{\circ}=\frac{v_m}{v_{r m}} \\ & \frac{1}{\sqrt{3}}=\frac{\text { Velocity of man }}{\text { Velocity of rain with respect to man }} \\ & \Rightarrow \quad v_{r m}=\frac{5}{1 / \sqrt{3}}=5 \sqrt{3} \mathrm{~km} / \mathrm{h}\end{aligned}$
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