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

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

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

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

answer is C.

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Detailed Solution

$y = a x - b x^{2}$

For height or $y$ to be maximum,

$\frac{d y}{d x} = 0$ or $a - 2 b x = 0$

or $x = \frac{a}{2 b}$

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

$(i i) {(\frac{d y}{d x})}_{x = 0} = a = \tan θ_{0}$ $(where θ_{0} = angle of projection)$

$∴$ $θ_{0} = \tan^{- 1} (a)$

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