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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 secαHorizontal component of velocity =vx=ucosθ Vertical component of velocity vy=usinθ−gtangle of velocity with horizontal =αtana=vyvx=(usinθ−gt)(4cosθ)t=u(sinθ−cosθtanα)gvy=ucosθtanαvx=ucosθSo the speed of the projectile =√((vx)2+(vy)2)=ucosθsecα

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