Answer
$h=1m$, $r=\sqrt 2m$ and $V=\frac{2\pi}{3}m^3$
Work Step by Step
Step 1. Using the Pythagorean Theorem, we have $h^2+r^2=3$ or $r^2=3-h^2$.
Step 2. The volume of the cone is given by $V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi (3-h^2) h=\frac{1}{3}\pi (3h-h^3)$
Step 3. $V'=\frac{1}{3}\pi (3-3h^2)$; let $V'=0$ to get $h=1m$; thus $r=\sqrt 2m$ and $V=\frac{1}{3}\pi (3-1^3)=\frac{2\pi}{3}m^3$
Step 4. Check $V''=\frac{1}{3}\pi (-6h)\lt0$ and the region is concave down with a maximum volume.
