Projectile motion: kinematics of a launch

Projectile motion describes the path of a body launched with initial velocity v₀ at angle θ to horizontal, under gravity g only. The trajectory is a perfect parabola if air resistance is neglected. This calculator solves the three fundamental variables — range R, max height h, time of flight t — plus impact velocity and landing angle.

Fundamental formulas

Decompose initial velocity into components:

• v_x = v₀ · cos(θ) — horizontal (constant, no force)
• v_y = v₀ · sin(θ) — vertical (decreases with g)

For launch on level ground (h₀ = 0):

• Range: R = v₀² · sin(2θ) / g
• Max height: h = v₀² · sin²(θ) / (2g)
• Time of flight: t = 2v₀ · sin(θ) / g
• Time to apex: t_apex = v₀ · sin(θ) / g (half of total time)
• Impact speed: |v| = v₀ (symmetric, no friction)

The optimal angle 45°

Formula R = v₀² · sin(2θ) / g is maximum when sin(2θ) = 1, i.e. 2θ = 90° → θ = 45°. On flat ground and without friction, 45° always gives maximum range regardless of v₀.

For angles symmetric about 45° (e.g., 30° and 60°), range is identical: sin(60°) = sin(120°). But height and flight time differ: 60° launches fly higher and stay longer in the air than 30°.

Range at 30° vs 45° vs 60° (v₀ = 20 m/s, g = 9.81)

Note: 30° and 60° give same range but very different heights and times.

Launch from initial height h₀ > 0

Launching from a height (roof, cliff) extends the time of flight: the projectile falls further before reaching ground. Full formula:

t = (v_y + √(v_y² + 2g·h₀)) / g

Range becomes R = v_x · t, greater than R(h₀=0).

Optimal angle for max range from h₀ > 0 is less than 45°:

θ_opt = arctan(v₀ / √(v₀² + 2g·h₀))

Example: v₀ = 20 m/s from h₀ = 10 m → θ_opt ≈ 39°. Higher elevation, lower optimal angle.

Practical examples

Example 1: kicked ball

A ball kicked at 25 m/s at 30°.

• R = 25² × sin(60°) / 9.81 = 625 × 0.866 / 9.81 = 55.2 m
• h = 25² × sin²(30°) / 19.62 = 7.96 m
• t = 2 × 25 × sin(30°) / 9.81 = 2.55 s

Example 2: volleyball serve

Speed 15 m/s, angle 8° (nearly horizontal), served from 2.2 m height.

• v_x = 14.85 m/s, v_y = 2.09 m/s
• Discriminant: 2.09² + 2 × 9.81 × 2.2 = 47.53
• t = (2.09 + √47.53) / 9.81 = 0.916 s
• R = 14.85 × 0.916 = 13.6 m (nearly court length)

Example 3: shot put (athletics)

Pro athlete: v₀ ≈ 13 m/s, angle ~38° (optimal given launch from 2.1 m height).

• R ≈ 22 m — world records ~23 m
• Optimal angle for elevated launches drops toward 38-40° (not 45°)

Example 4: compound bow arrow

v₀ = 55 m/s, angle 1.5° (nearly flat), h₀ = 1.5 m.

• Range ≈ 17 m before arrow hits ground
• Without angle, arrow drops visibly
• Archers compensate with inclined sight

Formula derivation

From Newton's second law with constant g and no other forces:

• a_x = 0 → v_x(t) = v₀·cos(θ)
• a_y = -g → v_y(t) = v₀·sin(θ) - g·t
• x(t) = v_x · t
• y(t) = h₀ + v_y·t − ½g·t²

Setting y(t) = 0 (ground impact) gives t; substituting into x(t) gives R. Max height at t_apex = v_y / g (when v_y=0).

Role of air resistance

The ideal model works well for dense objects at low speeds: cannonballs, arrows, pellets. For light or fast objects air drag dramatically changes the trajectory.

Real air effects

• Reduces range: a football at 30 m/s loses 10-20% distance vs vacuum
• Reduces max height
• Makes trajectory asymmetric: steeper descent than ascent
• Shifts optimal angle to 30-42° (not 45°)

Drag force F = ½·ρ·C_d·A·v², with C_d drag coefficient (≈0.47 sphere), ρ air density, A frontal area.

Secondary effects

• Magnus effect: spinning projectiles curve (topspin tennis ball falls faster)
• Wind: alters range and direction
• g variation: g = 9.83 at poles, 9.78 at equator
• Coriolis: for very long ranges (artillery), Earth's rotation deflects projectile

Applications of projectile motion

Sports

• Soccer: kick trajectories, crosses, free kicks
• Basketball: high-arc shots over defense
• Baseball: home runs fly 130-150 m
• Volleyball: serves and spikes
• Athletics: long jump, shot put, discus, javelin
• Golf: 300 m drives, dimpled ball effect

Military and historical

• Ancient artillery (catapults, trebuchets)
• Modern ballistics
• Artillery firing tables

Engineering

• Sprinklers (irrigation radius)
• Fire pumps (nozzle range)
• Physics animation in video games
• Pool splash design

Astronomy

• Satellite trajectories (limit: orbital velocity)
• Interplanetary probes (multi-body gravity)
• Atmospheric re-entries

Trajectory graph

Cartesian equation (eliminating t from x and y) is a parabola:

y = h₀ + x · tan(θ) − (g · x²) / (2 · v₀² · cos²(θ))

Negative x² coefficient, downward-opening parabola. Vertex at max height. X-axis intersections at 0 and R.

Symmetries and special cases

Complementary angles (θ + θ' = 90°)

θ = 30° and θ' = 60° give identical range (no drag, h₀=0). One low and fast, other high and slow. Same horizontal distance.

Vertical launch (θ = 90°)

Degenerate: v_x = 0, no horizontal range. Height: h_max = v₀²/(2g). Total time: t = 2v₀/g.

Horizontal launch (θ = 0°)

Projectile horizontal, falls by gravity. Fall time: t = √(2h₀/g). Range: R = v₀ · √(2h₀/g). Classic example: bullet fired from cliff.

45° + initial height

With h₀ > 0, optimal drops below 45°. For h₀ = 2 m, v₀ = 20 m/s → θ_opt ≈ 43.5°.

Energy conservation

Total mechanical energy is constant (closed system, no friction):

½m·v² + m·g·h = constant

• At launch: E_tot = ½m·v₀² + m·g·h₀
• At apex: E_tot = ½m·v_x² + m·g·h_max
• At impact: E_tot = ½m·v_impact²

So v_impact = √(v₀² + 2g·h₀). Without initial height, v_impact = v₀.

FAQ

What is the optimal angle for maximum range?

45° on level ground, no air resistance, zero initial height. With h₀ > 0, optimal angle is less than 45°. With air drag, real optimum is 30-42°.

Why do 30° and 60° give the same range?

Because sin(60°) = sin(120°) and range depends on sin(2θ). Complementary angles (θ and 90°-θ) share the same sin(2θ). But the high launch (60°) stays in the air much longer than the low (30°).

How does air resistance affect the motion?

Reduces range (10-40% depending on shape and speed), max height, and makes trajectory asymmetric. Shifts real optimal angle to 30-42°. Negligible for slow dense objects (cannonball).

Does the formula work on the Moon or other planets?

Yes, just change g. Moon: g = 1.62. Mars: g = 3.71. Jupiter: g ≈ 24.8. On the Moon, range at same v₀ is about 6× Earth range.

What is max range in vacuum vs in air?

Vacuum: 45°. Air: 30-42° depending on projectile. Example shot put record: ~38° because the ball is heavy but launched from h₀ = 2 m.

How do I calculate if the projectile clears an obstacle?

Compute y(x_obstacle) = h₀ + x·tan(θ) − g·x²/(2·v₀²cos²θ). If y > h_obstacle, clears it. Else hits.

Is impact speed equal to launch speed?

Only if h₀ = 0 and no friction. With h₀ > 0 (high launch), v_impact = √(v₀² + 2g·h₀) > v₀. With drag, v_impact is less.

History: from Galileo to modern ballistics

Aristotle (4th c. BC) believed projectiles moved straight until force ran out, then fell vertically — wrong but dominant for centuries.

Tartaglia (1537, Nova Scientia): first firing tables, hypothesized curved path.

Galileo (1638, Discorsi): mathematically proved trajectory is a parabola, combining uniform horizontal with uniformly accelerated vertical motion. Milestone of modern physics.

Newton (1687): second law F=ma unifies mechanics. Projectile motion becomes a special case.

Modern ballistics: fluid dynamics integration, artillery firing tables, Monte Carlo simulations.

Inverse calculation: angle to hit target

Given v₀ and target coordinates (x, y), two angles can reach it. Solving quadratic in tan(θ):

tan(θ) = (v₀² ± √(v₀⁴ − g·(g·x² + 2·y·v₀²))) / (g·x)

Real solution only if the square root argument ≥ 0. The low angle is faster (short time), high angle slower (high arc). Ballistics prefers the low angle for precision (less wind time).

Updated July 2026