RANGE GIVEN ANGLE AND SPEED

Start by splitting the initial velocity into components. The horizontal component vₓ = 20·cos(30°) ≈ 17.32 m/s stays constant for the entire flight because no horizontal force acts (air is off). The vertical component vy = 20·sin(30°) = 10 m/s decreases under gravity at a rate of g ≈ 9.807 m/s² per second. The ball is in the air until it returns to the same height it left. Setting y(t) = 0 gives two solutions: t = 0 (launch) and t = T = 2·vy/g = 2·10/9.807 ≈ 2.04 s. That is the time of flight. Range is horizontal velocity times time of flight: R = vₓ · T = 17.32 · 2.04 ≈ 35.3 m. Equivalently, the compact form R = v₀²·sin(2θ)/g = 400·sin(60°)/9.807 ≈ 35.3 m. Both routes give the same answer — use whichever you find more memorable. Note: 30° is not the optimum. Maximum range occurs at 45°, where sin(2θ) = sin(90°) = 1. Any angle between 0° and 90° gives a shorter range. Complementary angles — 30° and 60° — give exactly the same range, just via different trajectories.