A man running on a horizontal road at $8$ $k m / h$ finds the rain falling vertically. He increases his speed to $12$ $k m / h$ and finds that the drops make angle $30^{o}$ with the vertical. Find the speed and direction of the rain with respect to the road.

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Solution

This shows that the horizontal component of rain $= 8$ $k m / h r$ as the relative velocity of the rain has only a vertical component with respect the man.
$V_{x} = 8$ $k m / h r$
$2^{n d}$ case $\to {\to V}_{R M} = {\to V}_{R} - {\to V}_{M} = (8 ↑ - V_{y} ↑) - 12 ↑= - 4 ↑ - V_{y} ↑$
$tan 30^{o} = \frac{4}{V_{y}}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{4}{V_{y}}$
$V_{y} = 4 \sqrt{3}$ $k m / h r$
$\Rightarrow | V | = 4 \sqrt{7}$ $k m / h r$
Velocity of Rain $= 8$ $k m / h r ↑ - 4 \sqrt{3}$ $k m / h r ↑$
$\Rightarrow tan α = \frac{2}{\sqrt{3}}$ from vertical.

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