A metallic sphere with a radius of $\sqrt[z]{16}$ is recast into a cone, the lateral surface area of which is three times the area of the base. Find the height of the cone.
Let $R=\sqrt[3]{16}$ be the radius of the sphere, $V$ its volume, $r$ the radius of the base of the cone, $l$ the slant height, $h$ the height of the cone, $S$ the lateral surface area of the cone, and $s$ the area of its base. According to the problem,

$$
S=3 s, \text { or } \pi r l=3 \pi r^2 \text {, }
$$

$$
l=3 r, h^2=l^2-r^2=9 r^2-r^2=8 r^2, r^2=\frac{h^2}{8}
$$

According to the problem, the volume of the cone is equal to the volume of the sphere. Therefore,

$$
\begin{aligned}
& V=\frac{1}{3} \pi(\sqrt[3]{16})^3=\frac{1}{3} \cdot 16 \pi=\frac{16 \pi}{3} \\
& V=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi \cdot \frac{h^2}{8} \cdot h=\frac{\pi h^3}{24}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_acc13965d0370bfc03dcg-07.jpg?height=74&width=1378&top_left_y=409&top_left_x=0)

# Answer 

8.00