A girl standing on a road holds her umbrella | vedclass

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(C) Let $\vec{V}_R$ be the velocity of rain with respect to the ground and $\vec{V}_G$ be the velocity of the girl with respect to the ground.When the

Vedclass · Accepted Answer

$\frac{30}{\sqrt{2}}$

Vedclass · Answer

(C) Let $\vec{V}_R$ be the velocity of rain with respect to the ground and $\vec{V}_G$ be the velocity of the girl with respect to the ground.When the girl is standing,she holds the umbrella at $45^{\circ}$ with the vertical,which means the direction of rain with respect to the ground makes an angle of $45^{\circ}$ with the vertical.From the vector triangle,$	an 45^{\circ} = \frac{|\vec{V}_G|}{|\vec{V}_{R,G}|}$,where $\vec{V}_{R,G}$ is the velocity of rain with respect to the girl.When the girl starts running with velocity $\vec{V}_G = 15 \sqrt{2} \; kmh^{-1}$,the rain hits her head vertically. This means the relative velocity of rain with respect to the girl,$\vec{V}_{R,G} = \vec{V}_R - \vec{V}_G$,is vertical.From the geometry of the velocity triangle,$|\vec{V}_{R,G}| = \frac{|\vec{V}_G|}{	an 45^{\circ}}$.Since $	an 45^{\circ} = 1$,we have $|\vec{V}_{R,G}| = |\vec{V}_G| = 15 \sqrt{2} \; kmh^{-1}$.The speed of rain drops with respect to the moving girl is $|\vec{V}_{R,G}| = 15 \sqrt{2} \; kmh^{-1}$.Wait,checking the options provided: $15 \sqrt{2} \approx 21.21$. None of the options match. Let's re-evaluate: If the rain hits her vertically when she runs,then $\vec{V}_{R,G}$ is vertical. The horizontal component of $\vec{V}_R$ must be equal to the girl's speed $V_G = 15 \sqrt{2}$.Thus,$V_R \sin 45^{\circ} = 15 \sqrt{2} \Rightarrow V_R \frac{1}{\sqrt{2}} = 15 \sqrt{2} \Rightarrow V_R = 30 \; kmh^{-1}$.The vertical component of rain is $V_{R,G} = V_R \cos 45^{\circ} = 30 	imes \frac{1}{\sqrt{2}} = \frac{30}{\sqrt{2}} \; kmh^{-1}$.Thus,the correct option is $C$.

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