1. Velocity just before hitting the incline

The body is dropped from rest from height $H$ and hits the incline when it has descended to height $h$.

So the vertical distance fallen before impact is $H-h.$

Using $v^2 = 2g(H-h),$ we get the speed just before impact: $v = \sqrt{2g(H-h)}.$

Since it was falling freely, this velocity is vertically downward.

2. Velocity just after impact

The impact is perfectly elastic and we are told that after collision the direction of velocity becomes horizontal.

In a perfectly elastic impact with a fixed smooth plane, the speed remains unchanged, only direction changes appropriately.

Hence just after impact, the body moves horizontally with speed $u = \sqrt{2g(H-h)}.$

Its vertical component just after impact is zero.

3. Time after impact to reach the ground

At the instant after impact, the body is at height $h$ above the ground and has zero vertical velocity.

So its subsequent vertical motion is simply free fall from height $h$ starting with zero vertical speed.

Thus, $h = \frac{1}{2}gt^2.$ So the time after impact is $t_2 = \sqrt{\frac{2h}{g}}.$

4. Time before impact

The body falls freely from height $H$ to height $h$, i.e. through distance $H-h$.

Thus, $H-h = \frac{1}{2}gt_1^2,$ so $t_1 = \sqrt{\frac{2(H-h)}{g}}.$

5. Total time to reach the ground

Therefore, $T = t_1 + t_2 = \sqrt{\frac{2(H-h)}{g}} + \sqrt{\frac{2h}{g}}.$

Factor out $\sqrt{\frac{2}{g}}$: $T = \sqrt{\frac{2}{g}}\left(\sqrt{H-h}+\sqrt{h}\right).$

We must maximize $T$ with respect to $h$ for fixed $H$.

So maximize $f(h)=\sqrt{H-h}+\sqrt{h}.$

6. Maximization

Differentiate: $f'(h)= -\frac{1}{2\sqrt{H-h}}+\frac{1}{2\sqrt{h}}.$

Set $f'(h)=0$: $\frac{1}{\sqrt{h}}=\frac{1}{\sqrt{H-h}}.$ Hence, $\sqrt{h}=\sqrt{H-h}$ which gives $h=H-h.$ So, $H=2h.$ Therefore, $\frac{H}{h}=2.$

7. Comparison with stored answer

Our derived answer is $\boxed{2}.$

This matches the stored correct answer.