Projectile motion

After launch, and while air resistance is negligible, the only force on a projectile is its weight, so a⃗ = −g ĵ with ĵ pointing up and g = 9.81 m s⁻². Choose one axis along that acceleration and one across it, and the vector equation splits into a component with no acceleration and a component with a constant one. Integrate from a launch at speed v₀ and angle θ: vₓ = v₀ cos θ never changes, and v_y = v₀ sin θ − g t falls steadily. Integrate again: x = (v₀ cos θ) t and y = h + (v₀ sin θ) t − ½ g t² for a launch height h. A stone thrown at 16.0 m s⁻¹, 50.0° above the horizontal, keeps vₓ = 10.3 m s⁻¹ the whole way, while v_y starts at 12.3 m s⁻¹ and passes zero after 12.3/9.81 = 1.25 s. At that apex the stone still moves at 10.3 m s⁻¹, and its acceleration is still 9.81 m s⁻² downward. Only the vertical component of the velocity has vanished.

The x and y equations are independent in that neither acceleration touches the other, but they describe one object, so the t in each is the same number. That fixes the method: take the time from the component whose condition you know, almost always a height, then spend that time in the other. Throw a ball at 10.0 m s⁻¹, 30.0° above the horizontal, from a ledge 8.00 m above flat ground. Landing means y = 0, so 0 = 8.00 + 5.00t − 4.905t², with roots t = 1.88 s and t = −0.865 s. The negative root is not an error: it is when the same parabola, run backwards, would have crossed ground level, and the model does not apply before launch. So x = (8.66 m s⁻¹)(1.88 s) = 16.3 m. A target above the launch height can give two positive roots, the rising pass and the falling pass, and the question decides which one counts. What never works is a vertical time from one event paired with a horizontal time from another.

Time labels points on the path but does not shape it. Put the origin at the launch point, substitute t = x/(v₀ cos θ) into y(t), and you get the trajectory equation y = x tan θ − g x²/(2v₀² cos² θ). Close to the launcher the linear term wins, so the path leaves along the launch direction; the quadratic term bends it down with d²y/dx² = −g/vₓ², fixed by the horizontal speed alone. The slope is dy/dx = tan θ − g x/(v₀² cos² θ), and substituting x = (v₀ cos θ) t turns it into (v₀ sin θ − g t)/(v₀ cos θ) = v_y/vₓ: the velocity is tangent to the path, as it must be. Setting the slope to zero puts the apex at x = v₀² sin θ cos θ/g, which for the 16.0 m s⁻¹, 50.0° stone is 12.8 m out and, from the trajectory equation, 7.66 m up. This form answers questions that never mention time, such as whether a ball clears a wall at a given distance and whether it is still rising as it does.

When the landing height equals the launch height, y = 0 has the roots t = 0 and T = 2v₀ sin θ/g. The apex comes at T/2, at height H = v₀² sin² θ/(2g), and the range is R = vₓT = 2v₀² sin θ cos θ/g = v₀² sin 2θ/g. Because sin 2θ = sin(180° − 2θ) = sin 2(90° − θ), the angles θ and 90° − θ give the same range. At 24.0 m s⁻¹, launches at 35° and 55° both land 55.2 m away, but they do not share a flight: the 35° launch peaks at 9.66 m and lands after 2.81 s, the 55° launch peaks at 19.7 m and lands after 4.01 s. Two exact links tie the pair, T₁T₂ = 2R/g and H₁ + H₂ = v₀²/(2g). Setting dR/dθ = (2v₀²/g) cos 2θ to zero puts the maximum at 45°, Rₘₐₓ = v₀²/g = 58.7 m. That maximum is flat: 10° either side of it the range falls by only 6%, so near 45° the landing point hardly depends on the aim.

Replace 1/cos² θ with 1 + tan² θ and, writing a = g x²/(2v₀²), the trajectory equation becomes a quadratic in tan θ for any target (x, y): a tan² θ − x tan θ + (y + a) = 0. Two real roots mean two aims, so the level-ground pair was never a coincidence. The sum and product of the roots give tan(θ₁ + θ₂) = −x/y, which means θ₁ + θ₂ = 90° + α, where α is the target's elevation seen from the launcher: exactly 90° only when α = 0, more for a target above, less for one below. Where the discriminant vanishes the two aims merge, and those points trace the reach envelope y = v₀²/(2g) − g x²/(2v₀²), a parabola enclosing everything the launcher can hit. Setting y = −h gives the longest range from a height h, R = (v₀/g)√(v₀² + 2gh), reached at tan θ = v₀/√(v₀² + 2gh). From 10.0 m up at 15.0 m s⁻¹ that is 31.4 m at 36.2°, not at 45°.

Every result here assumes uniform g, flat ground and no air. The first two are safe for anything thrown or launched near the ground: g falls by 1% only about 32 km up, and Earth's curvature matters only over ranges of many kilometres. Drag is the assumption that fails in ordinary life. Compare quadratic drag, F = ½ρC_dAv², with air density ρ = 1.2 kg m⁻³, drag coefficient C_d and frontal area A = πd²/4, against the weight mg. A 7.26 kg shot 0.120 m across, released at 13.5 m s⁻¹ with C_d ≈ 0.47, feels 0.58 N of drag against 71 N of weight, under 1%, and the parabola is excellent. A 45.9 g, 42.7 mm golf ball leaving the club at 70 m s⁻¹ with C_d ≈ 0.25 feels 1.05 N against 0.45 N: drag is more than twice the weight, and the vacuum parabola is not even a first approximation. Drag bleeds horizontal speed all flight long, so the descent is steeper than the climb, the apex sits beyond the middle of the range, the best angle drops below 45°, and of a complementary pair the steeper launch, in the air longer, lands shorter.