Question

# The vertical height of the projectile at time t is given by y=4t−5t2 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 tan−1(4/3)Displacement in vertical direction at time t, y=4t−5t2 ⇒uy=dydt=4−10t Initial velocity at t=0 uy=4−0=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=tan−1(4/3)

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