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Question

A projectile is fired with a speed u at an angle θ with horizontal. Its speed when its direction of motion makes an angle 'α'with the horizontal is

A
u secθ cosα
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B
u secθ sinα
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C
u cosθ secα
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D
u sinθ secα
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Solution

The correct option is C u cosθ secα
Horizontal component of velocity =vx=ucosθ
Vertical component of velocity vy=usinθgt
angle of velocity with horizontal =α
tana=vyvx=(usinθgt)(4cosθ)
t=u(sinθcosθtanα)g
vy=ucosθtanα
vx=ucosθ
So the speed of the projectile =((vx)2+(vy)2)
=ucosθsecα

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