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Q.

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

$\frac{2 a^{2}}{b}, {Tan}^{- 1} (a)$

b

$\frac{a^{2}}{b}, {Tan}^{- 1} (2 b)$

c

$\frac{a^{2}}{4 b}, {Tan}^{- 1} (a)$

d

$\frac{b^{2}}{2 a}, {Tan}^{- 1} (b)$

answer is D.

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Detailed Solution

$C o m p a r i n g t h e g i v e n e q u a t i o n w i t h t h e s \tan d a r d e q u a t i o n y = x \tan θ - \frac{g s e c^{2} θ x^{2}}{2 u^{2}}, w e c a n w r i t e, Tan θ = a \Rightarrow θ = {Tan}^{- 1} (a), A l s o, b = \frac{g s e c^{2} θ}{2 u^{2}} \Rightarrow u^{2} = \frac{g (1 + a^{2})}{2 b} a n d a l s o \sin^{2} θ = \frac{a^{2}}{1 + a^{2}} T h e r e f o r e, H = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{a^{2}}{4 b} .$

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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 )