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Question

The vertical height of the projectile at time t is given by y=4t5t2 and the horizontal distance covered is given by x=3t where x and y are in m and t in s. What is the angle of projection with the horizontal?

A
tan1(3/5)
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B
tan1(4/5)
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C
tan1(4/3)
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D
tan1(3/4)
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Solution

The correct option is C tan1(4/3)
Displacement in vertical direction at time t,
y=4t5t2
uy=dydt=410t
Initial velocity at t=0
uy=40=4 m/s
Horizontal displacement at time t,
x=3t
vx=dxdt=3
Initial velocity at t=0
ux=3 m/s


Now, angle of projection with horizontal,
tanθ=uyux=43
θ=tan1(4/3)

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