Step 1: Choosing base radius as the variable:
Let the sphere have radius $r$ and center $O$. Let the cone have base radius $R$ and height $h$.
Since the apex and the base circle lie on the sphere, geometry gives $h=r+\sqrt{r^2-R^2}$ (the larger root is taken, since this gives the taller cone which can be the maximum).
Step 2: Volume as a function of R:
Volume of cone is $V=\dfrac{1}{3}\pi R^2 h$.
Substituting h:
\[ V(R)=\dfrac{1}{3}\pi R^2\left(r+\sqrt{r^2-R^2}\right), \quad 0<R<r \]
Step 3: Differentiating with respect to R:
Differentiate and simplify.
\[ \dfrac{dV}{dR}=\dfrac{\pi R}{3\sqrt{r^2-R^2}}\left(2r^2-3R^2+2r\sqrt{r^2-R^2}\right) \]
Set $\dfrac{dV}{dR}=0$. Since $R\neq0$, solve $2r\sqrt{r^2-R^2}=3R^2-2r^2$.
Step 4: Solving for R and finding h:
Square both sides and simplify.
\[ 4r^2(r^2-R^2)=(3R^2-2r^2)^2 \implies 9R^4=8r^2R^2 \implies R^2=\dfrac{8r^2}{9} \]
So $h=r+\sqrt{r^2-\dfrac{8r^2}{9}}=r+\dfrac{r}{3}=\dfrac{4r}{3}$. Checking the sign of the derivative on either side of this point confirms this is a maximum.
Final Answer:
Both methods confirm the same result for the maximum-volume cone.
\[ \boxed{h=\dfrac{4r}{3}} \]