The horizontal range and maximum height attained by a projectile are $R$ and $H$ , respectively. If a constant horizontal acceleration $a = \frac{g}{4}$ is imparted to the projectile due to wind, then its horizontal range and maximum height will be

(R + H), \frac{H}{2}

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(R + H), 2 H

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(R + 2 H), H

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(R + H), H

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Solution

The correct option is D $(R + H), H$
$H = \frac{u^{2} {sin}^{2} θ}{2 g}, R = \frac{u^{2} sin 2 θ}{g}$ Maximum height will be same because acceleration $a = g$ is in horizontal direction.
$R^{'} = u cos θ T + \frac{1}{2} a T^{2}$
Since time of flight, $T = \frac{2 u sin θ}{g}$
$∴ R^{'} = u cos θ (\frac{2 u sin θ}{g}) + \frac{1}{2} \frac{g}{4} {(\frac{2 u sin θ}{g})}^{2}$
$∴ R^{'} = R + \frac{1}{2} \frac{g}{4} {(\frac{2 u sin θ}{g})}^{2} = R + H$

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