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(C) Let $\vec{v}_{rg}$ be the velocity of rain with respect to the ground,$\vec{v}_{mg}$ be the velocity of the man with respect to the ground,and $\v

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$10\sqrt{3} \, km/h$

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(C) Let $\vec{v}_{rg}$ be the velocity of rain with respect to the ground,$\vec{v}_{mg}$ be the velocity of the man with respect to the ground,and $\vec{v}_{rm}$ be the velocity of the rain with respect to the man.
From the relative velocity equation: $\vec{v}_{rg} = \vec{v}_{rm} + \vec{v}_{mg}$,which implies $\vec{v}_{rm} = \vec{v}_{rg} - \vec{v}_{mg}$.
When the man is at rest,the rain falls at $30^\circ$ to the vertical. Let $v_r$ be the magnitude of the rain's velocity. Then $\vec{v}_{rg} = v_r \sin 30^\circ \hat{i} - v_r \cos 30^\circ \hat{j}$.
When the man runs at $10 \, km/h$,his velocity is $\vec{v}_{mg} = 10 \hat{i}$.
The velocity of rain relative to the man is $\vec{v}_{rm} = (v_r \sin 30^\circ - 10) \hat{i} - v_r \cos 30^\circ \hat{j}$.
Since the rain hits the man vertically,the horizontal component of $\vec{v}_{rm}$ must be zero:
$v_r \sin 30^\circ - 10 = 0 \implies v_r (1/2) = 10 \implies v_r = 20 \, km/h$.
The speed of raindrops with respect to the man is the vertical component: $v_{rm} = v_r \cos 30^\circ = 20 	imes (\sqrt{3}/2) = 10\sqrt{3} \, km/h$.

$A$ man standing on a road holds his umbrella at $30^\circ$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10 \, km/h$. He finds that raindrops are hitting his head vertically. The speed of raindrops with respect to the moving man will be:

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