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 vsinθ
b
velocity of particle at Q is vcotθ
c
time of flight from P to Q is (v/g)cosecθ
d
time of flight from P to Q is (v/g)secθ
answer is B.
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Detailed Solution
Initial velocity of particle in vector form can be written asvp=vcosθi^+vsinθj^Velocity of particle at any time t will bevQ=vcosθi^+(vsinθ−gt)j^Given that vP⊥vQ∴ vP⋅vQ=0or v2cos2θ+v2sin2θ−(vsinθ)gt=0or v2=(vsinθ)gt or t=vgcosecθSubstituting this value oft in Eq. (2) we get :vQ=vcosθi^+vsinθ−vsinθj^vQ=v2cos2θ+v2sin2θ+v2sin2θ−2v2=vcotθ