A particle is projected with an initial speed u from a point at height h above the  horizontal plane as shown in the fig. The maximum range on the horizontal plane is        $u\sqrt{\frac{u^2+ngh}{g^2}}$, then n is _____
 

Suppose the range R corresponding to the angle of projection  $\theta$
Then       $R=u\cos \theta t$……(1)
 $-h=\frac{1}{2}g{t}^2=u\sin \theta t--\left(2\right)$             
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 t2 should be unione. 
 $\therefore {(gh+u^2)}^2-g^2(R^2+h^2)=0$
 $\therefore R_{\max }=u\sqrt{\frac{u^2+2gh}{g}}$