A small object loops a vertical loop of radius $R$ in which a symmetrical section of $2\alpha$ has been removed. Find the maximum and minimum heights from which the object after loosing contact with the loop at point $A$ and flying through the air, will reach point $B$. Then the possible values of H are 

Let $v_0$ be the speed of the object at point $A$. Between $A$ and $B$ path of the object is a parabola, where

$AB=$range

or $2R \sin \alpha =\frac{2 v_0^2 \sin \alpha \cos \alpha }{g}$

So, $v_0^2=\frac{gR}{\cos \alpha } ...\left(1\right)$

Applying conservation of mechanical energy at $P$ and $A$ we get

$mgH=mgR\left(1+\cos \alpha \right)+\frac{1}{2} m{v}_0^2$

or $\frac{H}{R}=1+\cos \alpha +\frac{v_0^2}{2gR}$

but $\frac{v_0^2}{gR}=\frac{1}{\cos \alpha }$        (from equation $\left(1\right)$)

$\therefore \frac{H}{R}=1+\cos \alpha +\frac{1}{2 \cos \alpha }=k$ (say)

or $2 {\cos }^2 \alpha -2(k-1) \cos \alpha +1=0$

or ${\cos }^2 \alpha -(k-1) \cos \alpha +\frac{1}{2}=0$

or $\cos \alpha =\frac{1}{2}\left(k-1\pm \sqrt{{\left(k-1\right)}^2-2}\right)$

now ${\left(k-1\right)}^2-2\ge 0$ or $k-1\ge \sqrt{2}$

or $k\ge (1+\sqrt{2}) .....\left(2\right)$

On the other hand,

$\cos \alpha \le 1$

i.e., $\frac{1}{2}\left[(k-1)+\sqrt{{(k-1)}^2-2}\right]\le 1$

or $\left(k-1\right)+\sqrt{{\left(k-1\right)}^2-2} \le 2$

or $\sqrt{{(k-1)}^2-2}\le {\left[2-(k-1)\right]}^2$

or $4k \le 10$

or $k\le 2.5 ...\left(3\right)$

Hence from $\left(2\right)$ and $\left(3\right)$ we have

$1+\sqrt{2} \le k\le 2.5$

or $\left(1+\sqrt{2}\right)R \le H\le 2.5 R \left(as k=\frac{H}{R}\right)$

or $2.414 R\le H\le 2.5 R$