Let the height of the cone $=h$
and the radius of the cone $=r$
Given, radius of the sphere $=R$
Now, In $\triangle O P B$


$\Rightarrow \quad R^2=r^2+(h-R)^2$
$\Rightarrow \quad r^2=R^2-(h-R)^2$
$=(R+h-R)(R-h+R)$
$\Rightarrow \quad r^2=h(2 R-h)$
The volume of the cone is
$V=\frac{1}{3} \pi r^2 h$
$\Rightarrow \quad V=\frac{1}{3} \pi h(2 R-h) h$
$\Rightarrow \quad V=\frac{\pi}{3}\left(2 R h^2-h^3\right)$
Differentiating with $r$ to $h$
$\frac{d V}{d h}=\frac{\pi}{3}\left(4 R h-3 h^2\right)$
For maximum or minimum value of volume
$\frac{d V}{d h}=0$
$\Rightarrow \quad \frac{\pi}{3}\left(4 R h-3 h^2\right)=0$
$\Rightarrow \quad h(4 R-3 h)=0$
$\Rightarrow \quad h=0, h=\frac{4 R}{3} \quad$ (Not possible)
Now, $\quad \frac{d^2 V}{d h^2}=\frac{\pi}{3}(4 R-6 h)$
$\left(\frac{d^2 V}{d h^2}\right)_{\left(\text {at } h=\frac{4 R}{3}\right)}=\frac{\pi}{3}\left(4 R-6 \cdot \frac{4 R}{3}\right)$
$=\frac{\pi}{3}(4 R-8 R)=-\frac{4 \pi}{3} R \Rightarrow$ Negative
ie., Maximum
Hence, the height of the cone of maximum volume is $\left(\frac{4 R}{3}\right)$.