jmc

We draw and label a diagram as follows:

Let the center of the hemisphere be $O$, and let $A$ be a point on the circumference of the top circle of the cylinder. Since the cylinder is inscribed in the hemisphere, $A$ lies on the hemisphere as well, so $OA=5$. We drop a perpendicular from $A$ to the base of the hemisphere and let it intersect the base of the hemisphere at $B$. Since the cylinder is right and $AB$ is a height of the cylinder, $\angle OBA$ is a right angle, and $B$ lies on the circumference of the bottom circle of the cylinder. Thus, $OB$ is a radius of the cylinder, so $OB=2$. We have that $△OBA$ is right, so by the Pythagorean theorem, we have $$AB=\sqrt{O{A}^2-O{B}^2}=\sqrt{5^2-2^2}=\sqrt{21}.$$Thus, the height of the cylinder is $\boxed{\sqrt{21}}$.