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Question

# The trajectory of a projectile in a vertical plane is given by y=axâˆ’bx2, where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are

A
b22a,tan1(b)
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B
a2b,tan1(2a)
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C
a24b,tan1(a)
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D
2a2b,tan1(a)
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Solution

## The correct option is C a24b,tan−1(a)Trajectory of projectile in a vertical plane is given as y=ax−bx2y=ax⎛⎜ ⎜ ⎜ ⎜⎝1−x(ab)⎞⎟ ⎟ ⎟ ⎟⎠ ....(i) Compare (i) with y=xtanθ[1−xR] ⇒tanθ=a and R=ab ∴θ=tan−1(a) Now, For maximum height H, we have maximum height, H=u2sin2θ2g and Range R=u2sin2θg So, we have the ratio HR=u2sin2θ2gu2sin2θg ⇒ Rtanθ=4H ⇒ ab×a=4H⇒H=a24b

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