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Question

The trajectory of a projectile in a vertical plane is given by y=axbx2, 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,tan1(a)
Trajectory of projectile in a vertical plane is given as
y=axbx2y=ax⎜ ⎜ ⎜ ⎜1x(ab)⎟ ⎟ ⎟ ⎟ ....(i)
Compare (i) with y=xtanθ[1xR]
tanθ=a and R=ab
θ=tan1(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=4HH=a24b

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