Physics - Kinematics Question with Solution | TestHub

$T=\frac{2U_y}{g}$

$H=\frac{U_y^2}{2g}$

Here maximum height will be same as $H$ because there is no acceleration change in vertical direction.

and $R=U_x\times T$

When a horizontal acceleration is also given to the projectile $U_y,T$ and $H$ will remain unchanged while range will become

$R^{\text{'}}=U_x\times T+\frac{1}{2}a{T}^2$

$=R+\frac{1}{2}\frac{g}{4}\left(\frac{4U_y^2}{g^2}\right)=R+H \left[\because a=\frac{g}{4}\right]$

Therefore horizontal range & maximum height will be $\left(R+H\right) \& H$.