The volume of a cylinder is given by
![V=\pi hr^2](https://img.qammunity.org/2020/formulas/mathematics/high-school/v85ggrrpek9jmmwifjvy2ixg7kcx6z9ran.png)
Whereas the volume of a cone is given by
![V=(1)/(3)\pi hr^2](https://img.qammunity.org/2020/formulas/mathematics/high-school/eun4ntksb65j69xxfyrx4vodwxuun19120.png)
Since the cylinder has radius x and height y, its volume is
![V = \pi x^2y](https://img.qammunity.org/2020/formulas/mathematics/high-school/zmpjwufl9bq49021xliikry8gwj7ua3kte.png)
And since the cone has a radius 2x, its volume is
![V = \pi (2x)^2 h = 4\pi x^2h](https://img.qammunity.org/2020/formulas/mathematics/high-school/bbr3geqcipf7pb65t3whzdvqprl7tojxvh.png)
where h is the height of the cone.
We know that the two volumes are the same, so we can build the equation and solve it for h:
![\pi x^2y = 4\pi x^2h](https://img.qammunity.org/2020/formulas/mathematics/high-school/f1xq2wszuxl16fnwlc5jhufk01tjo6q3yc.png)
Simplify
from both sides:
![y = 4h](https://img.qammunity.org/2020/formulas/mathematics/high-school/g8bhk5ncq7zvoptuv71a93l9k2fexl1f8y.png)
Divide both sides by 4:
![h = (y)/(4)](https://img.qammunity.org/2020/formulas/mathematics/high-school/gt0a6e5ggn2m7sdun3p5f1rhn2nks24bt3.png)