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Question

A man is coming down an incline of angle 30. When he walks with speed 3m/s he has to keep his umbrella vertical w.r.t. ground to protect himself from rain. The actual speed of rain is 5 m/s. At what angle with vertical should he keep his umbrella when he is at rest so that he does not get drenched?


A

θ=57

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B

θ=37

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C

θ=30

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D

θ=45

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Solution

The correct option is B

θ=37


vman=23cos30^i23sin30^j

vm=3^i3^j

At this moment the rain appears to him as vertical.Let the magnitude of |vrm|=v

vrm=v^j

vrm=vrvm

vr=vrm+vm

Draw vector diagram

|vr|=5

Given now from it we know

vr=5 sin θ ^i5cos θ^j

vr=vrm+vm

5 sinθ^i=v^j+3^i3^j

(5sinθ=3) -------------(1)

5cosθ=(v+3)

From equation (1) we know sinθ=35

θ=37


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