A body starts falling freely from height $$H$$ hits an inclined plane in its path at height $$h$$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $$\frac{H}{h}$$ for which the body will take the maximum time to reach the ground is _____.

A body falls freely from height $$H$$, hits an inclined plane at height $$h$$, and after a perfectly elastic impact, its velocity becomes horizontal. We need to find $$\frac{H}{h}$$ for which the total time to reach the ground is maximized.

We start by finding the velocity at the inclined plane.

The body falls from height $$H$$ to height $$h$$, so it falls through a distance $$(H - h)$$. Using energy conservation (or kinematics):

$$ v = \sqrt{2g(H - h)} $$

Next, we analyze the motion after the perfectly elastic impact.

After the perfectly elastic impact, the direction of velocity changes to horizontal (magnitude remains the same). From height $$h$$ above the ground, the body undergoes projectile motion with:

- Initial horizontal velocity: $$v = \sqrt{2g(H-h)}$$

- Initial vertical velocity: 0 (horizontal direction)

Then, we calculate the total time.

Time for free fall from $$H$$ to $$h$$ (falling distance $$H - h$$):

$$ t_1 = \sqrt{\frac{2(H-h)}{g}} $$

Time for projectile fall from height $$h$$ to ground (vertical free fall from rest):

$$ t_2 = \sqrt{\frac{2h}{g}} $$

The total time is:

$$ T = t_1 + t_2 = \sqrt{\frac{2(H-h)}{g}} + \sqrt{\frac{2h}{g}} = \sqrt{\frac{2}{g}}\left(\sqrt{H-h} + \sqrt{h}\right) $$

To maximize the total time with respect to $$h$$, we consider:

We need to maximize $$f(h) = \sqrt{H - h} + \sqrt{h}$$ for $$0 \leq h \leq H$$.

Taking the derivative and setting it to zero:

$$ f'(h) = \frac{-1}{2\sqrt{H-h}} + \frac{1}{2\sqrt{h}} = 0 $$

$$ \frac{1}{2\sqrt{h}} = \frac{1}{2\sqrt{H-h}} $$

$$ \sqrt{h} = \sqrt{H - h} $$

$$ h = H - h $$

$$ 2h = H $$

$$ h = \frac{H}{2} $$

Finally, we find the ratio:

$$ \frac{H}{h} = \frac{H}{H/2} = 2 $$

The answer is $$\boxed{2}$$.