To solve the exercise let's remember that the volume of a cone is equal to
[tex]V=\frac{\pi r^2h}{3}=\frac{A_ch}{3}[/tex]
Where r is de radius and h is the height. Note that Ac is the circle's base of the cone. Solving the equation as a function of the height and replacing the information we have from the exercise
[tex]\begin{gathered} \frac{A_ch}{3}=V \\ A_ch=V3 \\ h=\frac{3V}{A_c}=\frac{3(30ft^3)}{15ft^2}=\frac{90ft^3)}{15ft^2}=6ft \\ \end{gathered}[/tex]
Thus, the height of the cone is 6 ft.The correct answer is D.
