The equation of trajectory of a projectile is given as $y = 2 x - \frac{x^{2}}{2}$ . The maximum height of projectile is (Symbols have usual meanings and SI unit)

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4 m

1 m

2 m

10 m

answer is C.

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Detailed Solution

$y = 2 x - \frac{xh}{2} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = 2 - \frac{2 x}{2} = 0$
When $y = h_{max} b \Rightarrow x = 2 m$
$\begin{array}{l} ∴ h_{max} = 2 \times 2 - \frac{2 \times 2}{2} \\ = 2 m \end{array}$

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