Correct option: B.$\sqrt{2 g H+\frac{R^2 g}{8 H}}$
b
(b)
$H=\frac{u^2 \sin ^2 \theta}{2 g} \Rightarrow \sin \theta=\sqrt{\frac{2 g H}{u^2}}$
$R=\frac{2 u^2 \sin \theta \cos \theta}{g}$
$R=\frac{2 u^2}{g} \sqrt{\frac{2 g H}{u^2}} \times \sqrt{1-\frac{2 g H}{u^2}}$
$R=\frac{2 u^2}{g} \sqrt{\frac{2 g H}{u^2}} \times \sqrt{\frac{u^2-2 g H}{u^2}}$
$\frac{g R}{2 \sqrt{2 g H}}=\sqrt{u^2-2 g H}$
Squaring both the sides,
$\frac{g R^2}{4 \times 2 g H}=u^2-2 g H$
$\Rightarrow u^2=2 g H+\frac{9 R^2}{8 H}$
$u=\sqrt{2 g H+\frac{g R^2}{8 H}}$