check the picture below, notice the base of the cone is really just a circle
thus [tex]\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\qquad
\begin{cases}
r=radius\\
h=height\\
----------\\
V=42\\
r=\sqrt{\frac{9}{\pi }}
\end{cases}\\\\
-----------------------------\\\\
42=\cfrac{\pi \left(\sqrt{ \frac{9}{\pi }} \right)^2h}{3}[/tex]
solve for "h"
