A girl standing on a road holds her umbrella | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $C$.

Similar Questions

$A$ train is moving towards the east and a car is moving towards the north,both with the same speed. The observed direction of the car to a passenger in the train is

$A$ man crosses a $320 \ m$ wide river perpendicular to the current in $4 \ minutes$. If in still water he can swim with a speed $5/3$ times that of the current,then the speed of the current,in $m/min$ is

$A$ man swimming downstream overtakes a float at a point $M$. After travelling a distance $D$,he turns back and passes the float at a distance of $D/2$ from the point $M$. The ratio of the speed of the swimmer with respect to still water $(v_s)$ to the speed of the river $(v_r)$ is:

$A$ boat crosses a river with a velocity of $8 \, km/h$ relative to the water. If the resultant velocity of the boat with respect to the ground is $10 \, km/h$,then the velocity of the river water is...........$km/h$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module