A particle is projected with an initial speed u form a point at height h above the horizontal plane as shown is the fig. Find the maximum range on the horizontal plane

The correct option is A

$u\sqrt{\frac{u^2+2gh}{g}}$

Suppose the range R corresponding to the angle of projection $\theta$

Then $R=ucos\theta t---\left(1\right)$
$-h=\frac{1}{2}g{t}^2=usin\theta t----\left(2\right)$
From (1)and (2)
$R^2+{(\frac{1}{2}g{t}^2-h)}^2=u^2{t}^2\Rightarrow \frac{1}{4}{g}^2{t}^4-(gh+u^2)t^2+(R^2+h^2)=0$
For the maximum range $t^2$ should be unione.
$\therefore (gh+u^2{)}^2-g^2(R^2+h^2)=0\therefore R_{max}=u\sqrt{\frac{u^2+2gh}{g}}$