Let the radius of the cone be $r$, the height be $h$, and the radius of the sphere be $R$. By the Pythagorean Theorem, we have that \[r^2+h^2=R^2=5^2=25.\]Thus, the equation is $r^2+h^2=25.$

The volume of the cone is given by $V=\frac13\pi r^2h,$ so we want to maximize $V$ subject to the constraint $r^2+h^2=25.$

Using a standard method of calculus, we set up Lagrange multipliers to maximize $V$ given the constraint. We wish to find the value of $r$ and $h$ that make the following equation true: \[\nabla V=\lambda\nabla C,\]where $\nabla V$ and $\nabla C$ are the gradients of $V$ and $C=r^2+h^2-25$, respectively. Taking the gradient of $V$ and $C$, we find that \[\nabla V=\left<\frac23\pi rh^2,\pi r^2h\right> \quad \text{and} \quad \nabla C=\left<2r,2h\right>.\]Then, our equation becomes \[\left<\frac23\pi rh^2,\pi r^2h\right>=\lambda\left<2r,2h\right>.\]Setting the corresponding components equal, we find that \[\frac{\frac23\pi rh^2}{2r}=\frac{\pi r^2h}{2h},\]so we can conclude that $\frac13 rh=\frac12 rh$ or that $r=0$. However, by the constraint $r^2+h^2=25,$ we know that $r\neq 0$. Therefore, this approach yields no solutions.

Instead, notice that we can rewrite the equation $r^2+h^2=25$ as $h=\sqrt{25-r^2}.$ Then, we can rewrite the volume of the cone as a function of $r$ alone: \[V=\frac13\pi r^2\sqrt{25-r^2}.\]Taking the derivative of $V$ with respect to $r$, we get \[V'(r)=\frac13\pi\left(2r\sqrt{25-r^2}-r^3\left(\frac{1}{\sqrt{25-r^2}}\right)\right).\]Setting this equal to $0$, we get \[2r\sqrt{25-r^2}=r^3\left(\frac{1}{\sqrt{25-r^2}}\right).\]This gives us two possibilities: either $2r\sqrt{25-r^2}=0$, which happens if $r=0$ (which we already labeled as extraneous), or we can divide both sides by $2r\sqrt{25-r^2}$ to give us \[1=\frac{r^2}{25-r^2}.\]Multiplying both sides by $25-r^2$ and rearranging terms, we get \[r^2=25-r^2,\]so $r^2=12.5$ and $r=\sqrt{12.5}=\frac{\sqrt{50}}{2}=\frac{5\sqrt{2}}{2}$. Then, $h=\sqrt{25-r^2}=\sqrt{25-\frac{50}{4}}=\sqrt{12.5}=\frac{\sqrt{50}}{2}=\frac{5\sqrt{2}}{2}$.

Therefore, the dimensions of the maximizing cone are $\boxed{\frac{5\sqrt{2}}{2}}$ cm and $\boxed{\frac{5\sqrt{2}}{2}}$ cm.

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