A body initially at rest and sliding along a friction-less track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to
 

If a body is moving on a frictionless surface, then its total mechanical energy remains conserved
According to the conservation of mechanical energy,
${\left(\text{TE}\right)}_{\text{initial}}={\left(\text{TE}\right)}_{\text{final}}$
$\Rightarrow {\left(\text{KE}\right)}_{\text{i}}+{\left(\text{PE}\right)}_{\text{i}}={\left(\text{KE}\right)}_{\text{f}}+{\left(\text{PE}\right)}_{\text{f}}$
$0+\text{mgh}=\frac{1}{2}{\text{mυ}}_{\text{A}}^2+0$
$\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}gh}=\frac{{\text{υ}}_{\text{A}}^2}{2}$
$\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}h}=\frac{{\text{υ}}_{\text{A}}^2}{2\text{\hspace{0.17em}g}}\dots . \left(i\right)$
In order to complete the vertical circle, the velocity of the body at point A should be
${\text{υ}}_{\text{A}}={\text{υ}}_{\min }=\sqrt{5\text{\hspace{0.17em}gR}}$
where, R is the radius of the body,
Here,  $\text{R}=\frac{\text{AB}}{\text{2}}=\frac{\text{D}}{\text{2}}$
$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_{\min }={\text{υ}}_{\text{A}}=\sqrt{\frac{5}{2}\text{gD}}$
Substituting the value of ${\text{υ}}_{\text{A}}$  in Equation (i), we get
$\text{h}=\frac{\left(\sqrt{\left(\frac{5}{2}\text{gD}\right)}\right)}{2\text{\hspace{0.17em}g}}=\frac{5\text{\hspace{0.17em}dD}}{2\times 2\text{\hspace{0.17em}g}}=\frac{5}{4}\text{D}$