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Question

A particle is launched from a horizontal plane with speed 10 m/s and angle of projection 60. The angular velocity of the particle as observed from the point of projection at the time of landing will be:

A
1 rad/sec
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B
2 rad/sec
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C
3 rad/sec
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D
4 rad/sec
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Solution

The correct option is A 1 rad/sec

When the particle lands, it will hit the ground with same speed of projection i.e., u.

With respect to line OP,

vparallel=vx=ucosθ

vperpendicular=vy=usinθ

So, angular velocity w.r.t point of projection,

ω=vperpendicularr

ω=usinθOP=usinθR

Here, R=u2sin2θg

ω=g usinθu2×2sinθcosθ

ω=g2ucosθ

Given, g=10 m/s2, θ=60,u=10 m/s2

ω=102×10cos60=1 rad/s

Hence, option (a) is the correct answer.

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