A man running on a horizontal road at 8kmph finds that rain is falling vertically. He increases the speed to 12kmphr and find that the drops make angle 30⁰ with the vertical

$\text{From }\Delta PQR\Rightarrow RQ=\left|{\overrightarrow{V}}_{m_1}\right|=\text{8km/hr}$

$\text{From }\Delta PSR\Rightarrow RS=\left|{\overrightarrow{V}}_{m_2}\right|=\text{12km/hr}$

$\text{From }\Delta \text{PQS}\Rightarrow \mathrm{t}\text{an}\left({\text{30}}^{\text{0}}\right)\text{=}\frac{\text{QR}}{\text{PQ}}$

$\frac{1}{\sqrt{3}}=\frac{4}{\left|{\overrightarrow{V}}_{R{M}_1}\right|}$

${\text{V}}_{\text{R}}\text{=}\sqrt{{\left({\text{V}}_{{\text{RM}}_{\text{1}}}\right)}^{\text{2}}\text{+}{\left({\text{V}}_{{\text{M}}_{\text{1}}}\right)}^{\text{2}}}\text{=}\sqrt{{\left(\text{8}\right)}^{\text{2}}\text{+}{\left(\text{4}\sqrt{\text{3}}\right)}^{\text{2}}}$

${\text{V}}_{\text{R}}\text{=}\sqrt{\text{112}}\text{ kmph}$

$\text{from }\Delta \text{PQR}\Rightarrow \text{Tan}\theta \text{=}\frac{\text{QR}}{\text{PQ}}\text{=}\frac{{|}_{{\text{M}}_{\text{1}}}|}{{|}_{{\text{RM}}_{\text{1}}}|}\text{=}\frac{{\text{V}}_{{\text{M}}_{\text{1}}}}{{\text{V}}_{{\text{RM}}_{\text{1}}}}\text{=}\frac{\text{8}}{\text{4}\sqrt{\text{3}}}\text{=}\frac{\text{2}}{\sqrt{\text{3}}}$

${\text{θ= tan}}^{\text{-1}}\frac{\text{2}}{\sqrt{\text{3}}}\text{with\hspace{0.17em}vertical\hspace{0.17em}}$