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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=uygt
v=ucos θ^i+(usinθgt)^j
If v is perpendicular to u then
v.u=0
u2cos2 θ+(u2sin2θusinθ gt)=0
usinθ gt=u2
t=ugsinθ
Since,
v=ucosθ^i+(usinθgt)^j......(i)
Using t=ugsinθ in equ. (i), we get
v=ucosθ^i+(usinθg×ugsinθ)^j
v=ucosθ^i+u(sin2θ1)sinθ^j
v=ucosθ^i+(ucos2θsinθ)^j
|v|=(ucosθ)2+u2cos4θsin2θ
|v|=(ucosθ)2(sin2θ+cos2θsin2 θ)
|v|=ucotθ

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