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

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$(R + 2 H), H$

$(R + H), H$

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

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

answer is D.

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Detailed Solution

$H = \frac{u^{2} \sin^{2} θ}{2 g}, R = \frac{u^{2} \sin 2 θ}{g}$

Maximum height will be same because acceleration
a = g/4 is in horizontal direction

$\begin{array}{l} R^{'} = u \cos θ T + \frac{1}{2} a T^{2} \\ = R + \frac{1}{2} \frac{g}{4} {(\frac{2 u \sin θ}{g})}^{2} = R + H \end{array}$

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