The trajectory of a projectile in a vertical plane is y = ax - 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
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a
b22a,tan-1(b)
b
a2b+tan-1(2b)
c
a24 b,tan-1(a)
d
2a2 b,tan-1(a)
answer is C.
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Detailed Solution
y=ax-bx2; for height or y to be maximum: dydx=0 or a-2bx=0 or x=a2b ymax=aa2 b-ba2 b2=a24 b dydxx-0=a=tanθ0, where θ0= angle of projection θ0=tan-1(a)