A man running at a speed of 5 text{ km/h} fin | vedclass

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(D) Let $\vec{v}_m$ be the velocity of the man,$\vec{v}_r$ be the velocity of the rain,and $\vec{v}_{rm}$ be the velocity of the rain with respect to

Vedclass · Accepted Answer

$5\sqrt{3} 	ext{ km/h}$

Vedclass · Answer

(D) Let $\vec{v}_m$ be the velocity of the man,$\vec{v}_r$ be the velocity of the rain,and $\vec{v}_{rm}$ be the velocity of the rain with respect to the man.Given,$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$.When the man is running,the rain appears to fall vertically,meaning $\vec{v}_{rm}$ is vertical.When the man stops,the rain falls at $60^{\circ}$ with the horizontal,meaning $\vec{v}_r$ makes an angle of $60^{\circ}$ with the horizontal.From the vector triangle,the angle between $\vec{v}_r$ and the vertical is $90^{\circ} - 60^{\circ} = 30^{\circ}$.In the right-angled triangle formed by $\vec{v}_m$,$\vec{v}_r$,and $\vec{v}_{rm}$:$	an(30^{\circ}) = \frac{|\vec{v}_m|}{|\vec{v}_{rm}|}$$\frac{1}{\sqrt{3}} = \frac{5}{|\vec{v}_{rm}|}$$|\vec{v}_{rm}| = 5\sqrt{3} 	ext{ km/h}$.

$A$ man running at a speed of $5 \text{ km/h}$ finds 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 the running man is

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