"v_0 = 125\\frac{\\text{m}}{\\text{s}}; \\alpha = 30\\degree; \\\\\nd = \\frac{v_0^2\\sin{2\\alpha}}{g} = \\frac{(125\\frac{\\text{m}}{\\text{s}})^2\\sin{(2\\cdot30\\degree)}}{9.81\\frac{\\text{m}}{\\text{s}^2}} \\approx 1379\\text{ m}."
For derivations, you may visit:
https://en.wikipedia.org/wiki/Range_of_a_projectile
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