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Question

The vertical height of the projectile at time t is given by y=4tt2 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 tan143
Given,
x=3t
y=4tt2
Initial velocity of projection along x-axis at t=0sec is,
vx=dxdt=3m/s
Initial velocity of projection along y-axis at t=0sec is
vy=dydt=42t
vy|t=0sec=4m/s
Velocity of the projection, v=vx^i+vy^j
v=3^i+4^j
Angle of projection, tanθ=vyvx
tanθ=43
θ=tan143
The correct option is C.

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