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Question

(B) Let the velocity of the man be $\vec{v}_m = 6 \hat{i} 	ext{ km/h}$ and the velocity of the rain be $\vec{v}_r = -6\sqrt{3} \hat{j} 	ext{ km/h}$.

Vedclass · Accepted Answer

$30^{\circ}$ with respect to vertical

Vedclass · Answer

(B) Let the velocity of the man be $\vec{v}_m = 6 \hat{i} 	ext{ km/h}$ and the velocity of the rain be $\vec{v}_r = -6\sqrt{3} \hat{j} 	ext{ km/h}$.To protect himself,the man must hold the umbrella in the direction of the relative velocity of the rain with respect to the man,$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$.$\vec{v}_{rm} = -6\sqrt{3} \hat{j} - 6 \hat{i} 	ext{ km/h}$.The angle $\alpha$ with respect to the vertical is given by $	an \alpha = \frac{|v_m|}{|v_r|} = \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}}$.Since $	an \alpha = \frac{1}{\sqrt{3}}$,we have $\alpha = 30^{\circ}$.Therefore,the man should hold his umbrella at an angle of $30^{\circ}$ with respect to the vertical.

$A$ man walking along a straight line with a velocity $6 \text{ km/h}$ encounters rain falling vertically down with a velocity $6\sqrt{3} \text{ km/h}$. At what angle should the man hold his umbrella to protect himself from the rain?

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