A body initially at rest and sliding along a frictionless 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 : 

\(\frac{5}{4}\) D

As track is frictionless, so total mechanical energy will remain constant 

mgh = \(\frac{1}{2}mv_L^2\)

h = \(\frac{v_L^2}{2g}\)

For completing the vertical circle, v_L \(\geq \) \(\sqrt{5gR}\)

h = \(\frac{5gR}{2g}\) = \(\frac{5}{4}\) D