1. **State the problem:** We have a tetrahedron (triangular pyramid) with a vertical height of 9 meters and a base that is an equilateral triangle with side length 7 meters. 2. **Recall the formula for the volume of a pyramid:** $$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$ 3. **Calculate the area of the equilateral triangular base:** The formula for the area of an equilateral triangle with side $a$ is $$A = \frac{\sqrt{3}}{4} a^2$$ Substitute $a = 7$: $$A = \frac{\sqrt{3}}{4} \times 7^2 = \frac{\sqrt{3}}{4} \times 49 = \frac{49\sqrt{3}}{4}$$ 4. **Calculate the volume:** $$V = \frac{1}{3} \times \frac{49\sqrt{3}}{4} \times 9 = \frac{49 \times 9 \times \sqrt{3}}{12} = \frac{441\sqrt{3}}{12} = \frac{147\sqrt{3}}{4}$$ 5. **Simplify if needed and approximate:** Approximate $\sqrt{3} \approx 1.732$: $$V \approx \frac{147 \times 1.732}{4} = \frac{254.604}{4} = 63.651$$ 6. **Final answer:** The volume of the tetrahedron is $$\boxed{\frac{147\sqrt{3}}{4} \text{ cubic meters} \approx 63.65 \text{ cubic meters}}$$