To find the volume of a cone, we use the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height of the cone.
Given:
- The radius \( r = 12 \) feet.
- The height \( h \) is half the length of the radius. Therefore,
\[ h = \frac{1}{2} \times r = \frac{1}{2} \times 12 = 6 \text{ feet}. \]
Now, we can substitute the radius and height into the volume formula:
\[ V = \frac{1}{3} \pi (12)^2 (6) \]
Calculating \( (12)^2 \):
\[ (12)^2 = 144 \]
Now, substituting this value into the volume formula:
\[ V = \frac{1}{3} \pi (144)(6) \]
Calculating \( 144 \times 6 \):
\[ 144 \times 6 = 864 \]
Now substituting this back into the volume formula:
\[ V = \frac{1}{3} \pi (864) \]
Using \( \pi = 3.14 \):
\[ V = \frac{1}{3} \times 3.14 \times 864 \]
Calculating \( \frac{1}{3} \times 864 \):
\[ \frac{1}{3} \times 864 = 288 \]
Now substituting this into the formula:
\[ V = 3.14 \times 288 \]
Calculating \( 3.14 \times 288 \):
\[ 3.14 \times 288 = 904.32 \]
Finally, rounding \( 904.32 \) to the nearest tenth:
\[ V \approx 904.3 \text{ cubic feet} \]
Thus, the volume of Erik’s cone is
\[ \boxed{904.3} \text{ cubic feet}. \]