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Question

A particle is projected from a point O in the horizontal surface OA with speed u and angle of projection θ. It just grazes the plane BC which makes an angle α with the horizontal. The time taken by the projectile to reach p from the instant of projection is
305873_e8defc7e9a59485fac00c593a74e2473.png

A
2usinθgcosθ
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B
usinθgcosα
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C
usin(θα)gcosα
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D
2usin(θα)gcosα
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Solution

The correct option is C usin(θα)gcosα
From figure, ux=ucosθ uy=usinθ
At point P, the particle just grazes the line BC at the instant t i.e VyVx=tanα
Also as ax=0 Vx=ux=ucosθ
Vy=Vxtanα=(ucosθ)tanα
y direction : Vy=uygt
ucosθtanα=usinθgt

OR t=u[sinθtanαcosθ]g=u[sinθcosαsinαcosθ]gcosα

t=usin(θα)gcosα

518093_305873_ans_53a24c48ded544eab93828b7f569abf7.png

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