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The equation of the projectile thrown from a point on the ground is $y = (x - x^{2} / 40) m$ . If $g = 10 m s^{- 2}$ , the maximum height reached is

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Detailed Solution

$y = x - (\frac{1}{40}) x^{2}$
In the form of $y = x \tan θ - (\frac{g}{2 u^{2} \cos^{2} θ}) x^{2}$
$\tan θ = 1, θ = 45^{\circ}$ and $\frac{2 u^{2} \cos^{2} θ}{g} = 40$

$h_{ma x} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{u^{2} \cos^{2} θ}{2 g} = 10 m$

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