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Physics Question on Motion in a plane

A man standing on the road has to hold his umbrella at 30 degrees from the vertical direction to keep the rain away. He thrown away the umbrella and starts running at 10 km h−1{10\, km \, h^{-1}}, he finds that the rain drops are hitting him vertically. The speed of the rain drops w.r.t. the road is

A

103\frac{10}{\sqrt{3}}

B

10

C

10310\sqrt{3}

D

20

Answer

20

Explanation

Solution

From equation of relative motion


V→RM+V→MG=V→RG\overrightarrow{V}_{RM} + \overrightarrow{V}_{MG} = \overrightarrow{V}_{RG}
∣V→MG∣sin⁡ γ=∣V→RG∣sin⁡ β=∣V→RG∣sin⁡(π−α)\frac{|\overrightarrow{V}_{MG} |}{\sin \, \gamma} = \frac{|\overrightarrow{V}_{RG} |}{\sin \, \beta} = \frac{|\overrightarrow{V}_{RG} |} {\sin( \pi - \alpha)}
According to question,
∣V→MG∣=10 km h−1,β=90∘|\overrightarrow{V}_{MG}|= {10 \, km \, h^{-1}} , \beta =90^\circ
∴  ∣V→MG∣sin⁡ 30∘=V→RGsin⁡ β⇒VRG=20km h−1\therefore \:\: \frac{|\overrightarrow{V}_{MG}|}{\sin \, 30^\circ} = \frac{\overrightarrow{V}_{RG}}{\sin \, \beta} \Rightarrow V_{RG} = {20 km \, h^{-1}}