Question

For a man walking towards east with velocity of 5 km/hr, rain appears to fall vertically down. When the man begins to walk towards west with a velocity of 3 km/hr, the rain appears to fall at an angle of $60{}^{\circ }$ to the vertical. Magnitude of velocity of rain, when the man stops walking is $\sqrt{\frac{139}{N}}$ km/hr, then $N$ is

Answer 1

Let velocity of rain is ${\overrightarrow{V}}_r=V_x\hat{i}-V_y\hat{j}$
Case - I :
$\overrightarrow{V_m}=5\hat{i}$
$\overrightarrow{V_{r,m}}=\overrightarrow{V_r}-\overrightarrow{V_m}=(V_x-5)\hat{i}-V_y\hat{j}$
Since, $V_x-5=0,V_x=5km/hr$

Case - II :
$\overrightarrow{V_m}=-3\hat{i}$
$\overrightarrow{V_{r,m}}=(V_x+3)\hat{i}-V_y\hat{j}=8\hat{i}-V_y\hat{j}$
As this velocity is inclined at ${60}^{\circ }$ to the vertical, we can say
$\tan {60}^{\circ }=\frac{8}{V_y}\Rightarrow V_y=\frac{8}{\sqrt{3}}$

So,
$V=\sqrt{V_x^2+V_y^2}=\sqrt{5^2+\frac{8^2}{3}}=\sqrt{\frac{139}{3}}$