🎯 Projectile Motion Calculator
Solve range, height, time and velocity — interactive animated trajectory, planet selector, and full step-by-step working
Projectile Motion Equations — Complete Formula Reference
Projectile motion separates into two independent components: horizontal (constant velocity) and vertical (constant acceleration due to gravity).
| Quantity | Formula | Notes |
|---|---|---|
| Horiz. velocity | vₓ = v₀cosθ | Constant throughout flight |
| Initial vert. velocity | vy₀ = v₀sinθ | At launch |
| Vert. velocity at t | vy = v₀sinθ − gt | Decreases with time |
| Horiz. position | x = v₀cosθ · t | Linear in time |
| Vert. position | y = v₀sinθ · t − ½gt² | Parabolic path |
| Time to peak | t_peak = v₀sinθ / g | Half of total flight time |
| Maximum height | H = (v₀sinθ)² / (2g) | At t = t_peak |
| Total flight time | T = 2v₀sinθ / g | Same-height launch only |
| Horizontal range | R = v₀²sin(2θ) / g | Same-height landing only |
| Speed at time t | v = √(vₓ² + vy²) | Pythagorean combination |
How Projectile Motion Works — The Two-Component Method
The key principle: horizontal and vertical motions are completely independent. Solve each separately, then combine.
Velocity stays constant: vₓ = v₀cosθ
Distance: x = vₓ × t
Velocity changes: vy = vy₀ − gt
Height: y = vy₀t − ½gt²
The range formula R = v₀²sin(2θ)/g is maximized when sin(2θ) = 1, i.e. when 2θ = 90°, giving θ = 45°. Maximum height is found using H = (v₀sinθ)²/(2g) — only the vertical velocity component contributes to height. Complementary angles (e.g. 30° and 60°) produce identical ranges but different heights and flight times.
| Angle θ | Range (m) | Max Height (m) | Time (s) |
|---|---|---|---|
| 15° | 20.39 | 1.34 | 1.05 |
| 30° | 35.35 | 5.10 | 2.04 |
| 45° ★ | 40.77 | 10.19 | 2.88 |
| 60° | 35.35 | 15.29 | 3.53 |
| 75° | 20.39 | 19.05 | 3.93 |
| 90° | 0 | 20.39 | 4.08 |
Values for v₀ = 20 m/s on Earth (g = 9.81 m/s²)
Why 45° Gives Maximum Range — The Mathematics Explained
The range formula is R = v₀²sin(2θ)/g. To maximize R we must maximize sin(2θ). Since the maximum value of any sine is 1, and sin(2θ)=1 when 2θ=90°, the optimal angle is θ = 45°, giving Rmax = v₀²/g.
| Initial Speed v₀ | R_max at 45° (m) | R_max at 45° (ft) |
|---|---|---|
| 5 m/s | 2.55 | 8.36 |
| 10 m/s | 10.19 | 33.43 |
| 20 m/s | 40.77 | 133.7 |
| 30 m/s | 91.74 | 300.8 |
| 50 m/s | 254.8 | 835.7 |
| 100 m/s | 1,019 | 3,343 |
Projectile Motion on the Moon and Other Planets
Gravity determines everything in projectile motion. On the Moon (g = 1.62 m/s²) the same kick sends a ball 6× farther. Use the planet selector above to compare.
| Body | g (m/s²) | Range (m) | Max H (m) | Time (s) |
|---|---|---|---|---|
| 🌍 Earth | 9.81 | 40.77 | 10.19 | 2.88 |
| 🌕 Moon | 1.62 | 246.9 | 61.73 | 17.42 |
| 🔴 Mars | 3.72 | 107.5 | 26.88 | 7.58 |
| 🟤 Jupiter | 24.79 | 16.14 | 4.03 | 1.14 |
| ⚫ Mercury | 3.70 | 108.1 | 27.03 | 7.62 |
| 🟡 Venus | 8.87 | 45.07 | 11.27 | 3.18 |
Values for v₀ = 20 m/s, θ = 45°
🏌️ During Apollo 14 (1971), astronaut Alan Shepard hit a golf ball on the Moon. With g = 1.62 m/s², even a moderate swing sent the ball vast distances in the low gravity — he estimated it went "miles and miles."
Projectile Motion in Real Life — Sports, Engineering & Nature
Projectile motion appears everywhere — from a football kick to a water fountain arc. Here are five real applications with worked numbers.
A penalty kick at v₀ = 28 m/s, θ = 16° travels 11 m horizontally.
x = v₀cosθ × t → t = 11 / (28 × cos16°) = 11 / 26.93 = 0.409 s
y = 28×sin16°×0.409 − ½×9.81×0.409² = 3.14 − 0.82 = 2.32 m — clears a 2.44 m crossbar only slightly. Launch angle matters enormously.
A free throw launched at v₀ = 7 m/s, θ = 51° from h₀ = 2 m must reach a basket at x = 4.6 m, y = 3.05 m.
t = 4.6 / (7 × cos51°) = 4.6 / 4.404 = 1.044 s
y = 2 + 7×sin51°×1.044 − ½×9.81×1.044² = 2 + 5.676 − 5.351 = 2.32 m — just below rim. The classic 51° "optimal" angle is validated by this calculation.
An athlete leaves the board at v₀ = 9.5 m/s, θ = 22°.
R = 9.5² × sin(44°) / 9.81 = 90.25 × 0.6947 / 9.81 = 6.39 m
World-class long jumpers achieve 8+ m because their takeoff speed exceeds 10 m/s. Every extra m/s at launch adds roughly 1.5 m to the range.
Decorative fountains are designed using projectile equations to hit a target landing point. A nozzle angled at 60° with v₀ = 4 m/s:
R = 4² × sin(120°) / 9.81 = 16 × 0.866 / 9.81 = 1.41 m
H = (4×sin60°)² / (2×9.81) = (3.464)² / 19.62 = 0.612 m — engineers use these numbers to route water precisely into basins.
This calculator uses simplified physics without air resistance — the standard assumption in introductory physics. Real ballistics software includes drag coefficients, wind, the Coriolis effect, and spin-induced deflection (Magnus effect). For a rifle bullet at v₀ = 900 m/s and θ = 0.5°, vacuum range = 900² × sin(1°) / 9.81 = 1,413 m, but actual range is far shorter due to aerodynamic drag.
Common Mistakes in Projectile Motion Problems
These five errors account for the majority of wrong answers in projectile motion problems:
Correct: H = (v₀sinθ)² / (2g) → only the vertical component lifts the projectile
For v₀=20, θ=30°: Wrong gives H=20.4 m, correct gives H=5.1 m
Correct: Always split first — vₓ = v₀cosθ (horizontal) and vy₀ = v₀sinθ (vertical) — then apply kinematic equations to each direction independently
Correct: y = vy₀t − ½gt² → gravity decelerates upward motion (g is magnitude; the minus sign accounts for direction)
Correct: t_peak = v₀sinθ/g is time to reach maximum height; total flight T = 2t_peak (for same-height launch). Using t_peak gives half the correct range.
Correct: R = v₀²sin(2θ)/g → the argument is 2θ, not θ. At 45°: sin(45°) = 0.707 vs sin(90°) = 1.0 — the wrong formula gives 29% less range.