Question

To a man walking at the rate of $3$ km/h the rain appears to fall vertically downwards. When he increases his speed to $6$ km /h it appears to meet him at an angle of ${45}^0$ with vertical. Find the speed of rain.

Answer 1

Let, $i$ and $j$ be unit vector in horizontal & vertical.

Then velocity of rain $V_r=ai+bj$

Speed of rain= $\sqrt{a^2+b^2}$

(1) When relative velocity of rain w.r.t max is vertical

$V_{rm}=\overline{V}r-\overline{V}m\Rightarrow V_m=3\overline{j}\Rightarrow V_{rm}=(a-3)i+bj$

$V_{rm}$ is vertical $a-3=0\Rightarrow a=3$

(2) When relative velocity is at ${45}^o$ $V_m=bi$

$V_{rm}=(a-b)i-bj\Rightarrow -3i+bj$

Since $\tan \theta =\frac{b}{-33}\Rightarrow \tan {45}^o=1\Rightarrow b=-3$

$\therefore$ Speed=$\sqrt{3^2+3^2}=3\sqrt{2}$ .