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Question

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 A ugsinθ
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=uygt
v=ucosθ^i+(usinθgt)^j
If v is perpendicular to u then
v.u=0
u2cos2θ+(u2sin2θusinθgt)=0
usinθgt=u2
t=ugsinθ

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