m/s
Please enter a valid initial velocity.
Launch speed in metres per second
degrees
Enter angle between 0 and 90.
Angle above horizontal. 45° = maximum range
m
Enter a valid height (0 or greater).
Height above landing surface. Leave 0 for ground-level
s
Enter a positive time value.
Find x, y position at this specific time
Gravitational acceleration
Horizontal Range
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Sources & Methodology
Trajectory equations verified against HyperPhysics, Khan Academy Physics, and NASA planetary gravity data.
HyperPhysics — Trajectory of a Projectile
Georgia State University reference for projectile trajectory equations, the parabolic path formula y(x), and derivation from kinematic equations.
Khan Academy — 2D Projectile Motion
Curriculum reference for position equations x(t) and y(t), independence of horizontal and vertical motion, and worked examples.
NASA — Planetary Fact Sheet
Source for gravitational acceleration values on Moon (1.62), Mars (3.72), Venus (8.87), and Jupiter (24.79) m/s².
Formulas used: x(t) = v₀cosθ × t. y(t) = h + v₀sinθ × t − ½g × t². Parabola: y(x) = h + x·tanθ − gx²/(2v₀²cos²θ). Max height H = h + (v₀sinθ)²/(2g). Flight time T (elevated): [v₀sinθ + √((v₀sinθ)² + 2gh)] / g. Range R = v₀cosθ × T.
⏱ Last reviewed: March 2026
How to Calculate the Trajectory of a Projectile
The trajectory of a projectile is the path it follows through space under gravity alone, with no air resistance. This path is always a parabola. Two sets of equations govern projectile motion — one for each dimension — which can be combined into a single trajectory equation.
Position Equations
x(t) = v₀ × cos(θ) × t
y(t) = h + v₀ × sin(θ) × t − ½ × g × t²
x(t) = horizontal position at time t (m)
y(t) = vertical position (height) at time t (m)
h = launch height above landing surface (m)
v₀ = initial speed (m/s) • θ = launch angle • g = 9.81 m/s²
Example at t = 1 s (v₀ = 20 m/s, θ = 45°, h = 0):
x = 20 × cos(45°) × 1 = 14.14 m
y = 0 + 20 × sin(45°) × 1 − ½ × 9.81 × 1² = 14.14 − 4.91 = 9.24 m
y(t) = vertical position (height) at time t (m)
h = launch height above landing surface (m)
v₀ = initial speed (m/s) • θ = launch angle • g = 9.81 m/s²
Example at t = 1 s (v₀ = 20 m/s, θ = 45°, h = 0):
x = 20 × cos(45°) × 1 = 14.14 m
y = 0 + 20 × sin(45°) × 1 − ½ × 9.81 × 1² = 14.14 − 4.91 = 9.24 m
The Parabolic Trajectory Equation
y(x) = h + x × tan(θ) − g × x² / (2 × v₀² × cos²(θ))
This eliminates time t to give height y directly as a function of horizontal position x.
The negative x² term confirms the downward-opening parabola shape.
Example (v₀ = 20 m/s, θ = 45°, at x = 20 m):
y = 0 + 20×1 − 9.81×400/(2×400×1) = 20 − 9.81×400/800 = 20 − 4.905 = 15.10 m
The negative x² term confirms the downward-opening parabola shape.
Example (v₀ = 20 m/s, θ = 45°, at x = 20 m):
y = 0 + 20×1 − 9.81×400/(2×400×1) = 20 − 9.81×400/800 = 20 − 4.905 = 15.10 m
Trajectory at Different Launch Angles (v₀ = 20 m/s, Earth gravity)
| Angle | Max Height | Range | Flight Time | Shape |
|---|---|---|---|---|
| 15° | 1.36 m | 20.36 m | 1.05 s | Flat, wide |
| 30° | 5.10 m | 35.35 m | 2.04 s | Shallow arc |
| 45° | 10.19 m | 40.77 m | 2.88 s | Balanced — max range |
| 60° | 15.29 m | 35.35 m | 3.53 s | Tall, narrow |
| 75° | 19.04 m | 20.36 m | 3.94 s | Very tall arc |
| 90° | 20.39 m | 0 m | 4.08 s | Straight up |
💡 Complementary angles produce equal range: 15° and 75° both give 20.36 m. 30° and 60° both give 35.35 m. The lower angle arrives faster and flatter; the higher angle arrives slower with a steeper descent. Athletes and engineers choose based on which trajectory shape best fits their need — a basketball arc requires a high angle, while a long-range shot benefits from a lower one.
Velocity Components Along the Trajectory
The horizontal velocity component vx = v₀cosθ remains constant throughout flight because no horizontal force acts on the projectile (ignoring air resistance). The vertical component vy = v₀sinθ − g×t decreases linearly, reaching zero at peak height, then becoming negative on the way down. The total speed at any point is v = √(vx² + vy²).
Real-World Applications
- Sports analytics: Ball trajectory analysis in basketball, soccer, tennis, and golf uses these equations to optimise shot angles, spin effects, and ideal release points.
- Military ballistics: Artillery and mortar firing solutions use trajectory calculations to account for range, elevation, and gravity — with corrections added for air resistance and wind.
- Aerospace engineering: Rocket trajectory calculations during the ballistic phase of flight use the same parabolic equations as a first approximation before atmospheric drag models are applied.
- Video game physics: Game engines implement projectile motion using these exact equations to simulate realistic bullet, grenade, and ball trajectories in real time.
- Civil engineering: Water jet trajectories from nozzles, spillways, and fountains are designed using parabolic trajectory equations to predict where water will land.
Frequently Asked Questions
The parabolic trajectory equation is y(x) = h + x·tan(θ) − g·x² / (2·v₀²·cos²(θ)). This gives height y at any horizontal distance x. For position at a specific time t: x(t) = v₀cos(θ)·t and y(t) = h + v₀sin(θ)·t − ½g·t². Both are derived from Newton’s equations of motion.
x(t) = v₀·cos(θ)·t and y(t) = h + v₀·sin(θ)·t − 0.5·g·t². For v₀=20 m/s, θ=45°, at t=2 s: x = 20×0.707×2 = 28.28 m, y = 0 + 20×0.707×2 − 0.5×9.81×4 = 28.28 − 19.62 = 8.66 m. At t = total flight time, y returns to 0 (ground level).
The trajectory is a parabola — a symmetric, U-shaped curve opening downward. This results from combining constant horizontal velocity with linearly increasing vertical velocity (due to gravity). Mathematically, a parabola is any quadratic curve of the form y = ax² + bx + c, and the projectile trajectory equation fits this form exactly with a = −g/(2v₀²cos²θ).
Maximum height H = h + (v₀×sinθ)² / (2g). It occurs at time t_peak = v₀×sinθ / g. For v₀=30 m/s, θ=60°: t_peak = 30×sin(60)/9.81 = 25.98/9.81 = 2.65 s. H = (30×sin60°)²/(2×9.81) = 675/19.62 = 34.4 m above the launch point.
For ground-level launch: R = v₀²×sin(2θ)/g. For launch from height h: first find flight time T = [v₀sinθ + √((v₀sinθ)² + 2gh)] / g, then R = v₀×cosθ×T. The projectile lands when y(t) = 0. From a 10 m cliff at 20 m/s, 45°: T = 4.20 s and R = 20×0.707×4.20 = 59.4 m.
A higher launch point shifts the entire parabola upward and extends the flight time and range. The shape of the arc is identical, but it starts higher and ends lower (at ground level further away). A projectile launched at 20 m/s, 45° from ground reaches 40.77 m range; from a 10 m cliff the same launch gives about 59.4 m. The extra height becomes extra flight time, which multiplies horizontal distance.
Horizontal velocity vx = v₀×cosθ is constant throughout. Vertical velocity vy = v₀×sinθ − g×t decreases linearly, is zero at peak height, then increases downward. Total speed = √(vx²+vy²). At landing on the same level, speed equals v₀ because kinetic energy is conserved. At peak, speed = vx = v₀×cosθ (minimum speed point).
Complementary angles (adding to 90°, like 30° and 60°) give the same horizontal range but completely different trajectory shapes. The 30° trajectory is low, flat, and fast; the 60° trajectory is tall, narrow, and slow. They land at the same point but via different paths. This is because sin(2θ) = sin(2(90°−θ)), so R = v₀²×sin(2θ)/g gives equal ranges.
Air resistance deforms the ideal parabola. The actual trajectory is shorter and asymmetric — the descent is steeper and shorter than the ascent. The optimal launch angle for maximum range drops below 45° (typically 30°–40° depending on shape and speed). For most physics problems and slow-moving dense objects, ignoring air resistance gives results within a few percent of reality. For bullets, baseballs, and fast-moving objects, computational drag models are necessary.
Time of flight gives a single number: the total airborne duration T. Trajectory analysis gives the complete path — x and y at every moment, the parabolic equation y(x), position at any specified time t, velocity components throughout, and the full arc shape. Trajectory is more comprehensive and is used when you need to know where the projectile is at intermediate points, not just when it lands.
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