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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. Find the magnitude of final velocity when it is perpendicular to the initial velocity.

A
ucotθ
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B
utanθ
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C
usinθ
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D
ucosθ
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Solution

## The correct option is A ucotθ→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=ucosθ (∵ Horizontal component of velocity always remains constant) vy=uy−gt →v=ucos θ^i+(usinθ−gt)^j If →v is perpendicular to →u then →v.→u=0 ⇒u2cos2 θ+(u2sin2θ−usinθ gt)=0 ⇒usinθ gt=u2 t=ugsinθ Since, →v=ucosθ^i+(usinθ−gt)^j......(i) Using t=ugsinθ in equ. (i), we get →v=ucosθ^i+(usinθ−g×ugsinθ)^j →v=ucosθ^i+u(sin2θ−1)sinθ^j ⇒→v=ucosθ^i+(−ucos2θsinθ)^j |→v|=√(ucosθ)2+u2cos4θsin2θ |→v|=√(ucosθ)2(sin2θ+cos2θsin2 θ) |→v|=ucotθ

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