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Question

# The vertical height of the projectile at time t is given by y=4tâˆ’t2 and the horizontal distance covered given by x=3t.What is the angle of projection with horizontal?

A
tan135
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B
tan145
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C
tan143
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D
tan134
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Solution

## The correct option is C tan−143Given,x=3ty=4t−t2Initial velocity of projection along x-axis at t=0sec is,vx=dxdt=3m/sInitial velocity of projection along y-axis at t=0sec isvy=dydt=4−2tvy|t=0sec=4m/sVelocity of the projection, →v=vx^i+vy^j→v=3^i+4^jAngle of projection, tanθ=vyvxtanθ=43θ=tan−143The correct option is C.

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