Question

A man walking with a speed of $3km/h$ finds the rain drops falling vertically downwards. When the man increases his speed to $6km{h}^{-1}$ he find that the rain drops are falling making an angle of ${30}^{\circ }$ with the vertical. Find the speed of the rain drops (in $km/h)$.

Answer 1

Let $\hat{i}$ and $\hat{j}$ be unit vector in horizontal and vertical respectively.

Let, velocity of rain, $v_r=a\hat{i}+b\hat{j}$

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

Case-I

When relative velocity of rain with respect to man is vertical.

$V_{rms}=\overrightarrow{V_r}-\overrightarrow{V_m}{V}_m=3\hat{i}{V}_{rms}=(a-3)\hat{i}+b\hat{j}$

Since, $V_{rms}$ is vertical

$a-3=0\Rightarrow a=3$

Case II

When relative velocity is at $30°$

$\overrightarrow{V_m}=6\hat{i}{V}_{rms}=(a-6)\hat{i}+b\hat{j}=3\hat{i}+b\hat{j}$

Since, $tan\theta =\frac{b}{-3}tan30°=\frac{b}{-3}$

$b=\frac{1}{\sqrt{3}}\times (-3)b=-\sqrt{3}|b|=\sqrt{3}$

The therefore, speed:-

$=\sqrt{3^2+(\sqrt{3}{)}^2}=\sqrt{(9+3)}=\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$