Conditioned For Maximum Time!

The body initially falls $H-h$ from a zero vertical velocity, which takes a time of $\sqrt{\frac{2}{g}(H-h)}.$ Then immediately after the collision, it once again has a zero vertical component of velocity, so it takes a time of $\sqrt{\frac{2}{g}h}$ to fall the remaining $h.$

The total time is $\sqrt{\frac{2}{g}(H-h)} + \sqrt{\frac{2}{g}h} = \sqrt{\frac{2H}{g}} \left(\sqrt{1-\frac{h}{H}} + \sqrt{\frac{h}{H}}\right)$ which, for a given $H,$ is maximized when $\sqrt{1-\frac{h}{H}} + \sqrt{\frac{h}{H}}$ is maximized, at $\frac{h}{H} = \boxed{\frac{1}{2}}$

The collision will convert all the vertical velocity to horizontal velocity. Since we are dealing with time we can consider that after the collision the vertical velocity again became 0

For Time before collision

(H-h) = 0T + $\frac {1 }{2}gT^2$

Therefore T = $\sqrt {\frac {2 (H - h)}{g}}$

After collision,

h = 0t + $\frac {1 }{2}gt^2$

=> t = $\sqrt {\frac {2 (h)}{g}}$

Therefore total time = $\sqrt {\frac {2H}{g}}(\sqrt {1 - \frac {h}{H}} + \sqrt {\frac {h}{H}})$

Now time will be maximum when $\frac {h}{H} = \frac {1}{2} = \boxed {0.5}$