Let the radius of the cone be $R$ and its height be $h$. Let $x$ be the radius and $y$ be the height of the cylinder inscribed in the cone. The semi-vertical angle of the cone is $\alpha$, so:

$$\tan \alpha = \frac{R}{h}$$

From the geometry of the figure, using similarity of triangles $\Delta QEC \sim \Delta AOC$, we have:

$$\frac{y}{h} = \frac{R-x}{R}$$

Solving for $y$, we get:

$$y = \frac{h(R-x)}{R}$$

The volume $V$ of the cylinder is given by $V = \pi x^2 y$. Substituting the expression for $y$, we get:

$$V(x) = \pi x^2 \left[ \frac{h(R-x)}{R} \right] = \frac{\pi h}{R} (Rx^2 - x^3)$$

Differentiating $V$ with respect to $x$, we get:

$$V'(x) = \frac{\pi h}{R} (2Rx - 3x^2)$$

For maxima or minima, $V'(x) = 0$:

$$x(2R - 3x) = 0$$

This gives $x = 0$ or $x = \frac{2R}{3}$. Differentiating again:

$$V''(x) = \frac{\pi h}{R} (2R - 6x)$$

At $x = \frac{2R}{3}$:

$$V''(x) = \frac{\pi h}{R} (2R - 4R) = -2\pi h < 0$$

Hence, $V$ is maximum at $x = \frac{2R}{3}$.

The height of the cylinder is:

$$y = \frac{h\left(R - \frac{2R}{3}\right)}{R} = \frac{h}{3}$$

This shows the height is one-third that of the cone. The greatest volume is:

$$V = \pi \left( \frac{2R}{3} \right)^2 \left( \frac{h}{3} \right) = \frac{4\pi R^2 h}{27}$$

Substituting $R = h \tan \alpha$, we get:

$$V = \frac{4\pi (h \tan \alpha)^2 h}{27} = \frac{4}{27}\pi h^3 \tan^2 \alpha$$

∴ The height of the cylinder is $\frac{h}{3}$ and the greatest volume is $\frac{4}{27}\pi h^3 \tan^2 \alpha$.