A particle is projected from the ground at an angle of θ with the horizontal with an initial speed of u. Time after which velocity vector of the projectile is perpendicular to the initial velocity is
A
ugsinθ
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B
ugcosθ
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C
2ugsinθ
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D
2utanθ
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Solution
The correct option is Augsinθ →u=ux^i+uy^j →u=ucosθ^i+usinθ^j Let v be the velocity which is perpendicular to u At any time t, vx=ux (∴ Horizontal component of velocity always remains constant) vy=uy−gt →v=ucosθ^i+(usinθ−gt)^j If →v is perpendicular to →u then →v.→u=0 ⇒u2cos2θ+(u2sin2θ−usinθgt)=0 ⇒usinθgt=u2 t=ugsinθ