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

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

$\frac{a^{2}}{b} + \tan^{- 1} (2 b)$

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

answer is C.

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Detailed Solution

$y = ax - {bx}^{2}; for height or y to be maximum: \frac{dy}{dx} = 0 or a - 2 bx = 0 or x = \frac{a}{2 b} y_{\max} = a (\frac{a}{2 b}) - b {(\frac{a}{2 b})}^{2} = \frac{a^{2}}{4 b} {(\frac{dy}{dx})}_{x - 0} = a = {tanθ}_{0}, where θ_{0} = angle of projection θ_{0} = \tan^{- 1} (a)$