The path of projectile is given by the equation y = ax - bx2 ,where ‘a’ and ‘b’ are constants and x and y are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively. ( 2014 - E )
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a
2a2b,Tan-1(a)
b
b22a,Tan-1(b)
c
a2b,Tan-1(2b)
d
a24b,Tan-1(a)
answer is D.
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Detailed Solution
Comparing the given equation with the standard equation y= x tan θ - g sec2θ x22u2, we can write , Tanθ=a⇒θ=Tan-1(a), Also , b = gsec2θ2u2 ⇒u2=g(1+a2)2b and also sin2θ = a21+a2 Therefore , H = u2sin2θ2g=a24b .