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Question

# A particle is projected from a point P with a velocity v at an angle θ with horizontal. At a certain point Q it moves at right angles to its initial direction. Then

A
velocity of particle at Q is vsinθ
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B
velocity of particle at Q is vcotθ
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C
time of flight from P to Q is vg cosec θ
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D
time of flight from P to Q is vgsecθ
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Solution

## The correct options are C velocity of particle at Q is vcotθ D time of flight from P to Q is vg cosec θLet net velocity at Q is vq, it makes an (θ−90) with horizontal as it is vertical to vq and vp makes θ with horizontal .Now as there is no acceleration along x and air resistance is ignored , horizontal component of velocity remains unchanged .i.e vpcosθ = vqcos(θ−90) vq = vpcotθ = vcotθ ..... (i) initial vertical velocity +vsinθFinal vertical component of velocity = vqsin(θ−90) =−vcotθ×cosθ = −v(cosθ)2/sinθ Using laws of motion :−v(cosθ)2sinθ=vsinθ−gtt=vcosecθ/g .... (ii) (Time of flight from P to Q )

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