A cone has a volume of $12288\pi$ cubic inches and the vertex angle of the vertical cross section is 60 degrees. What is the height of the cone? Express your answer as a decimal to the nearest tenth. [asy]

import markers;
size(150);
import geometry;
draw(scale(1,.2)*arc((0,0),1,0,180),dashed);
draw(scale(1,.2)*arc((0,0),1,180,360));
draw((-1,0)--(0,sqrt(3))--(1,0));

//draw(arc(ellipse((2.5,0),1,0.2),0,180),dashed);
draw(shift((2.5,0))*scale(1,.2)*arc((0,0),1,0,180),dashed);
draw((1.5,0)--(2.5,sqrt(3))--(3.5,0)--cycle);

//line a = line((2.5,sqrt(3)),(1.5,0));
//line b = line((2.5,sqrt(3)),(3.5,0));
//markangle("$60^{\circ}$",radius=15,a,b);
//markangle("$60^{\circ}$",radius=15,(1.5,0),(2.5,sqrt(3)),(1.5,0));
markangle(Label("$60^{\circ}$"),(1.5,0),(2.5,sqrt(3)),(3.5,0),radius=15);
//markangle(Label("$60^{\circ}$"),(1.5,0),origin,(0,1),radius=20);
[/asy]
The cross section of the cone is an equilateral triangle. The ratio of the base to the height of an equilateral triangle is 1 to $\sqrt{3}/2$. In terms of the radius, $r$, the base is $2r$ and the height is $2r\sqrt{3}/2$, or $r\sqrt{3}$. Since we know the volume of the cone, we can use the volume formula and solve the equation  \[(1/3) \times \pi \times r^2 \times r\sqrt{3} = 12288\pi\]  for $r$. Dividing both sides of the equation by $\pi$ gives  $(1/3)r^3\sqrt{3} = 12288$. Tripling both sides, we get $r^3\sqrt{3} = 36,\!864$. Now, we want $r\sqrt{3},$ so we multiply both sides by $3$ to get $r^3\cdot(\sqrt{3})^3 = (r\sqrt{3})^3 = 36,\!864 \cdot 3 = 110,\!592.$ Taking the cube root of both sides, we get $r\sqrt{3} = \boxed{48.0}.$
48.0
Level 5
Geometry
Problem: A cone has a volume of $12288\pi$ cubic inches and the vertex angle of the vertical cross section is 60 degrees. What is the height of the cone? Express your answer as a decimal to the nearest tenth. [asy]

import markers;
size(150);
import geometry;
draw(scale(1,.2)*arc((0,0),1,0,180),dashed);
draw(scale(1,.2)*arc((0,0),1,180,360));
draw((-1,0)--(0,sqrt(3))--(1,0));

//draw(arc(ellipse((2.5,0),1,0.2),0,180),dashed);
draw(shift((2.5,0))*scale(1,.2)*arc((0,0),1,0,180),dashed);
draw((1.5,0)--(2.5,sqrt(3))--(3.5,0)--cycle);

//line a = line((2.5,sqrt(3)),(1.5,0));
//line b = line((2.5,sqrt(3)),(3.5,0));
//markangle("$60^{\circ}$",radius=15,a,b);
//markangle("$60^{\circ}$",radius=15,(1.5,0),(2.5,sqrt(3)),(1.5,0));
markangle(Label("$60^{\circ}$"),(1.5,0),(2.5,sqrt(3)),(3.5,0),radius=15);
//markangle(Label("$60^{\circ}$"),(1.5,0),origin,(0,1),radius=20);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the height of the cone, we start with the formula for the volume of a cone, which is given by V = (1/3)πr²h, where r is the radius of the base and h is the height. We know the volume is 12288π cubic inches, so we can set up the equation:

12288π = (1/3)πr²h.

We can simplify this by dividing both sides by π:

12288 = (1/3)r²h.

Next, we multiply both sides by 3 to eliminate the fraction:

36864 = r²h.

Now, we need to relate the radius r to the height h using the vertex angle of the cone. The vertex angle is 60 degrees, which means that in the vertical cross-section, the cone forms an isosceles triangle. The angle at the vertex is split into two 30-degree angles.

Using trigonometry, we can express the radius in terms of the height. In a right triangle formed by the height, the radius, and the slant height, we have:

tan(30°) = r/h.

Since tan(30°) = 1/√3, we can write:

r = h * (1/√3).

Now we substitute this expression for r back into the volume equation:

36864 = (h * (1/√3))² * h.

This simplifies to:

36864 = (h²/3) * h,
36864 = h³/3.

Multiplying both sides by 3 gives:

110592 = h³.

Taking the cube root of both sides, we find:

h = ∛(110592).

Calculating the cube root, we find:

h ≈ 48.0.

Thus, the height of the cone is 48.0 inches.<|END_OF_LATENT|> The final answer is $48.0$.