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Q.

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,

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a

velocity of particle at Q is vsin⁡θ

b

velocity of particle at Q is vcot⁡θ

c

time of flight from P to Q is (v/g)cosec⁡θ

d

time of flight from P to Q is (v/g)sec⁡θ

answer is B.

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Detailed Solution

Initial velocity of particle in vector form can be written asvp=vcos⁡θi^+vsin⁡θj^Velocity of particle at any time t will bevQ=vcos⁡θi^+(vsin⁡θ−gt)j^Given that vP⊥vQ∴ vP⋅vQ=0or  v2cos2⁡θ+v2sin2⁡θ−(vsin⁡θ)gt=0or  v2=(vsin⁡θ)gt  or  t=vgcosec⁡θSubstituting this value oft in Eq. (2) we get :vQ=vcos⁡θi^+vsin⁡θ−vsin⁡θj^vQ=v2cos2⁡θ+v2sin2⁡θ+v2sin2⁡θ−2v2=vcot⁡θ
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