A person standing on a road has to hold his umbrella at ${60}^{°}$ with the vertical to keep the rain away. He throws the umbrella and starts running at $20 m{s}^{-1}$. He finds that rain drops are falling on him vertically. Find the speed of the rain drops with respect to 

1. The road 2. The moving person

Given $\theta ={60}^{°}$ and velocity of person ${\overrightarrow{V}}_P=\overrightarrow{OA} =20 m{s}^{-1}$.  

${\overrightarrow{V}}_{r/p}=\overrightarrow{OB}$= velocity of rain w.r.t. person.

${\overrightarrow{V}}_r=\overrightarrow{OC}$ =velocity of rain w.r.t .earth

Values of ${\overrightarrow{V}}_r$ and ${\overrightarrow{V}}_{r/p}$ can be obtained by using simple trigonometric relations

1. Speed of rain drops w.r.t. earth ${\overrightarrow{V}}_r =\overrightarrow{OC}$ from $∆OCB,\frac{CB}{OC}=\sin {60}^{°}$

$\Rightarrow OC=\frac{CB}{\sin {60}^{°}}=\frac{20}{\sqrt{3/2}}=\frac{40}{\sqrt{3}}=\frac{40\sqrt{3}}{3}m{s}^{-1}$

2. Speed of rain w.r.t. the person ${\overrightarrow{V}}_{r/p}=\overrightarrow{O}B$ From $∆OCM,\frac{OB}{CB}=\cot {60}^{°}$

$\Rightarrow OB=CB\cot {60}^{°}=\frac{20}{\sqrt{3}}=\frac{20\sqrt{3}}{3}m{s}^{-1}$