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Question

Physics Question on Motion in a plane

A man is walking due east at the rate of 2 km/h. The rain appears to him to come down vertically at the rate of 2 km/h. The actual velocity and direction of rainfall with the vertical respectively are

A

22km/h,4502\sqrt{2}\,km/h,{{45}^{0}}

B

12km/h,300\frac{1}{\sqrt{2}}\,km/h,{{30}^{0}}

C

2km/h,002\,km/h,{{0}^{0}}

D

1km/h,9001\,km/h,{{90}^{0}}

Answer

22km/h,4502\sqrt{2}\,km/h,{{45}^{0}}

Explanation

Solution

Velocity of man, vm=2km/h{{v}_{m}}=2\,km/h Velocity of rain, vr=2km/h{{v}_{r}}=2\,km/h Then actual velocity vrm=vm2+vr2{{v}_{rm}}=\sqrt{v_{m}^{2}+v_{r}^{2}} =(2)2+(2)2=\sqrt{{{(2)}^{2}}+{{(2)}^{2}}} =4+4=22km/h=\sqrt{4+4}=2\sqrt{2}\,km/h The direction of rainfall with the vertical sinθ=vrvrm\sin \theta =\frac{{{v}_{r}}}{{{v}_{rm}}} =222=12=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}} θ=sin1(12)\theta ={{\sin }^{-1}}\left( \frac{1}{\sqrt{2}} \right) θ=45o\theta ={{45}^{o}}