The problem combines fluid dynamics (Torricelli's Law) and projectile motion.

First, determine the exit velocity $v$ of the water from the hole using Torricelli's Law. The velocity of efflux from a hole at a depth $h$ is equivalent to the speed of an object falling freely from that height.

$$v = \sqrt{2gh}$$

Next, analyze the projectile motion of the water stream. The stream is launched horizontally from a height of $(H-h)$ above the floor. The time of flight $t$ can be found from the vertical motion equation, where the initial vertical velocity is zero.

$$H-h = \frac{1}{2}gt^2$$

Solving for time $t$:

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

The horizontal range $R$ is the product of the constant horizontal velocity $v$ and the time of flight $t$.

$$R = v \cdot t$$

Substituting the expressions for $v$ and $t$:

$$R = \sqrt{2gh} \cdot \sqrt{\frac{2(H-h)}{g}}$$ $$R = \sqrt{4h(H-h)}$$ $$R = 2\sqrt{h(H-h)}$$