The horizontal range and maximum height attained by a projectile...

T = \dfrac{2{U}_{y}}{g}$</p><p>$H = {U}_y^{2}2g$</p><p>Here maximum height will be same as$H$because there is no acceleration change in vertical direction.</p><p>and$R = {U}_{x} \times T$</p><p>When a horizontal acceleration is also given to the projectile${U}_{y}, T and H$will remain unchanged while range will become</p><p>$R^{'} = {U}_{x} \times T + \dfrac{1}{2}aT^{2}$</p><p>$= R + \dfrac{1}{2}\dfrac{g}{4} \left(\dfrac{4{U}_y^{2}}{g2}\right) =R + H \left[∵ a = \dfrac{g}{4}\right]</p><p>Therefore horizontal range & maximum height will be \left( R + H\right) & H.