so hmm check the picture below
the radius "r" is half the diameter, meaning, the diameter is 2r long
now, if the height "h" is twice "d" or 2d, then that means h = 2(2r)
thus [tex]\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\qquad
\begin{cases}
h=height\\
r=radius\\
------\\
h=2(2r)
\end{cases}\implies V=\cfrac{\pi r^2[2(2r)]}{3}
\\\\\\
V=\cfrac{4\pi r^3}{3}[/tex]
