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The trajectory of a projectile in a vertical plane is $y = a x - b x^{2},$ 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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$\frac{b^{2}}{2 a}, \tan^{- 1} (b)$

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

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

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

answer is C.

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Detailed Solution

$y = ax - {bx}^{2}$

For height or y to be maximum

$\frac{dy}{dx} = 0 or a - 2 bx = 0 or x = \frac{a}{2 b}$

$(i) y_{max} = a (\frac{a}{2 b}) - b {(\frac{a}{2 b})}^{2} = \frac{a^{2}}{4 b}$

$(ii) \frac{dy}{dx} = a - 2 bx$

an initial position x = 0

${\frac{dy}{dx}|}_{x = 0} = a = tanθ$

$θ = \tan^{- 1} (a)$

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