A man standing on a road has to hold his umbr | vedclass

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$(	ext{A})$ Let $\vec{v}_{r,g}$ be the velocity of rain with respect to the ground, $\vec{v}_{m,g}$ be the velocity of the man with respect to the gr

Vedclass · Accepted Answer

$20 \,km/h$

Vedclass · Answer

$(	ext{A})$ Let $\vec{v}_{r,g}$ be the velocity of rain with respect to the ground, $\vec{v}_{m,g}$ be the velocity of the man with respect to the ground, and $\vec{v}_{r,m}$ be the velocity of rain with respect to the man.When the man is at rest, the rain falls at $30^{\circ}$ with the vertical. Thus, the horizontal component of $\vec{v}_{r,g}$ is $v_{r,g} \sin 30^{\circ}$ and the vertical component is $v_{r,g} \cos 30^{\circ}$.When the man runs at $v_{m,g} = 10 \,km/h$, the rain appears to fall vertically. This means the horizontal component of the relative velocity $\vec{v}_{r,m} = \vec{v}_{r,g} - \vec{v}_{m,g}$ must be zero.Therefore, the horizontal component of $\vec{v}_{r,g}$ must be equal to the velocity of the man:$v_{r,g} \sin 30^{\circ} = v_{m,g}$$v_{r,g} 	imes (1/2) = 10 \,km/h$$v_{r,g} = 20 \,km/h$.

$A$ man standing on a road has to hold 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 actual speed of raindrops is:

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