Projectile Motion Calculator

Compute range, max height, time of flight, and velocity components from launch speed and angle. Assumes launch from ground level with no air resistance.

🎯 Kinematics📐 Parabolic Path🌍 g = 9.8 m/s²
Initial Speed (v₀)
Speed unit
Launch Angle (θ) in degrees
Gravity (g) m/s²
💡 For maximum range on flat ground, use 45°. For max height, use 90°.
⚠️ Please enter valid positive numbers. Angle must be 0–90°.
Horizontal Range
—
Max Height
—
Time of Flight
—
seconds
Impact Speed
—
m/s

What Is Projectile Motion?

Projectile motion describes the curved path of an object launched into the air and subject only to gravity (and, in the idealized model, no air resistance). What makes projectile motion elegant is the principle of independence of motion: the horizontal and vertical components of the trajectory are completely independent of each other and can be analyzed separately.

Horizontally, with no air resistance, there is no force acting, so the horizontal velocity (v_x = v₀·cos θ) remains constant throughout the flight. Vertically, gravity acts continuously, causing the object to accelerate downward at g = 9.8 m/s². The vertical velocity starts at v_y₀ = v₀·sin θ, decreases to zero at the peak, then increases again until landing.

The combination of constant horizontal motion and uniformly accelerated vertical motion produces a parabolic trajectory — the characteristic arc shape of thrown balls, launched projectiles, and water fountains.

Key relationships for launch from ground level: the range R = v₀²·sin(2θ)/g is maximized at 45°, and complementary angles (e.g., 30° and 60°) produce the same range. The maximum height H = v₀²·sin²θ/(2g) is maximized at 90° (straight up). The time of flight T = 2v₀·sin θ/g is the total duration before landing.

Real projectiles deviate from this ideal due to air resistance, spin effects (Magnus force), and curvature of Earth over long ranges — but for most physics problems and many engineering applications, the ideal parabola model is excellent.

Formula Reference Table

QuantityFormulaNotes
Horizontal velocityvx = v₀·cos θConstant throughout flight
Initial vertical velocityvy₀ = v₀·sin θDecreases due to gravity
Range (flat ground)R = v₀²·sin(2θ) / gMax at θ = 45°
Max heightH = v₀²·sin²θ / (2g)Height at peak of arc
Time of flightT = 2v₀·sin θ / gTotal air time
Height at time ty = vy₀·t − ½g·t²Vertical position
Horizontal positionx = vx·tHorizontal displacement
Horizontal cliff launchT = √(2h/g)From height h, horizontal launch

3 Worked Examples

Example 1
Soccer Kick — Optimal Angle

A ball is kicked at 25 m/s at a 45° angle. Find the range and time of flight.

  • Range: R = v₀²·sin(90°)/g = 25² × 1 / 9.8 = 625/9.8 = 63.8 m
  • Time of flight: T = 2 × 25 × sin(45°)/9.8 = 2 × 25 × 0.707/9.8 = 3.61 s
  • Max height: H = 25² × sin²(45°)/(2 × 9.8) = 625 × 0.5/19.6 = 15.9 m
✓ Range = 63.8 m | Flight time = 3.61 s | Max height = 15.9 m
Example 2
Javelin Throw — Non-45° Angle

A javelin is released at 28 m/s at 35°. Find range, max height, and impact speed.

  • Components: vx = 28 cos 35° = 22.94 m/s; vy₀ = 28 sin 35° = 16.06 m/s
  • Range: R = 28² × sin(70°)/9.8 = 784 × 0.9397/9.8 = 75.2 m
  • Max height: H = 16.06²/(2 × 9.8) = 258/19.6 = 13.2 m
  • Impact speed: vx = 22.94 m/s (horizontal); vy_impact = −vy₀ = −16.06 m/s → v = √(22.94² + 16.06²) = 28 m/s
✓ Range = 75.2 m | Max height = 13.2 m | Lands at 28 m/s
Example 3
Cliff Launch — Horizontal Projectile

A ball rolls off a 78.4 m cliff at 15 m/s horizontally. Where does it land and how fast?

  • Fall time: T = √(2h/g) = √(2 × 78.4/9.8) = √16 = 4 seconds
  • Horizontal range: x = vx × T = 15 × 4 = 60 m from base
  • Vertical speed at impact: vy = g × T = 9.8 × 4 = 39.2 m/s
  • Total impact speed: v = √(15² + 39.2²) = √(225 + 1537) = 41.9 m/s
✓ Lands 60 m from base after 4 s, at 41.9 m/s (150.8 km/h)

Real-World Applications

⚽
Sports Science
Coaches optimize launch angles for soccer kicks, javelin throws, and basketball shots. The ideal free-throw arc (~52°) ensures a steeper entry angle into the basket, giving more margin for error.
💧
Water Fountains & Irrigation
Engineers use projectile equations to design fountain arcs and sprinkler patterns, calculating nozzle angles and pressures to achieve specific coverage shapes and distances.
🎮
Video Game Physics
Every physics-based game engine (Unreal, Unity) implements projectile motion. Grenade trajectories, arrow arcs, and ball physics all use the same v₀, θ, and g framework.
🌋
Volcanology
Scientists use projectile models to predict where volcanic bombs and pyroclastic ejecta will land given eruption velocity, estimating evacuation zones around active volcanoes.
🚀
Ballistics
Artillery targeting uses projectile equations with corrections for air resistance, Earth's curvature (Coriolis effect), and muzzle velocity to calculate shell trajectories over long distances.

Common Mistakes to Avoid

⚠️
Using degrees without converting to radians in code

Math.sin() and Math.cos() in JavaScript (and most programming languages) expect radians. Multiply degrees by π/180 before using trig functions, or your answers will be wildly wrong.

⚠️
Forgetting the ½ in H = v₀²sin²θ/(2g)

The maximum height formula comes from v_y² = v_y0² − 2gH at the peak (where v_y = 0). Solving gives H = v_y0²/(2g). The factor of 2 in the denominator is essential.

⚠️
Applying ground-level formulas to cliff launches

R = v₀²sin(2θ)/g only applies when launch and landing heights are equal. From a cliff, the horizontal range is longer (more time to fall), requiring the full kinematic equations.

⚠️
Ignoring that impact speed equals launch speed (in ideal case)

Without air resistance, the projectile returns to the same height with the same speed. The direction changes (angle below horizontal = angle above at launch), but the magnitude is identical. Students sometimes recalculate instead of using this symmetry.

⚠️
Thinking 45° always maximizes range in real life

The 45° optimal angle assumes level ground and no air resistance. With air drag, the optimal angle drops to ~38–42°. Javelin throwers throw closer to 30–35° for this reason. On inclines, the optimal angle shifts further.

Frequently Asked Questions

Why does a 45° angle give maximum range on flat ground? ▾
Range R = v₀²sin(2θ)/g is maximized when sin(2θ) = 1, which occurs when 2θ = 90°, so θ = 45°. At this angle, you optimally split v₀ between horizontal and vertical — enough vertical velocity for extended hang time, enough horizontal to cover ground. At 44° or 46°, sin(2θ) is only slightly less than 1, so the effect is gradual near the optimum.
Why do 30° and 60° launch angles give the same range? ▾
Because R = v₀²sin(2θ)/g, and sin(2×30°) = sin(60°) = sin(120°) = sin(2×60°). In general, any two complementary angles (θ and 90°−θ) give sin(2θ) = sin(180°−2θ), producing identical ranges. However, the trajectories differ: the 60° shot goes much higher but stays in the air longer.
Does the mass of the projectile affect its range or height? ▾
In the ideal model without air resistance, no — mass cancels out. All objects launched at the same angle and speed land at the same place regardless of mass (Galileo's famous finding). With air resistance, heavier objects are less affected by drag relative to gravity, so they travel farther — explaining why a shot put travels farther than a ping pong ball given the same launch conditions.
How does the calculator handle a horizontal cliff launch (θ = 0°)? ▾
Use "Launch from Height" mode. For a horizontal launch from height h with no vertical component: T = √(2h/g), Range = v₀ × T, and impact speed = √(v₀² + (gT)²). These are just the standard equations with vy₀ = 0.
At what point is vertical velocity zero? ▾
At the peak (maximum height). The time to reach the peak is t_peak = vy₀/g = v₀·sinθ/g, exactly half the total flight time. At this moment, vertical velocity = 0 and horizontal velocity = vx = v₀·cosθ (still unchanged). The speed at the peak equals the horizontal velocity component.
How does air resistance change projectile motion? ▾
Air resistance (drag) acts opposite to velocity, reducing both horizontal and vertical speed continuously. It makes the trajectory asymmetric (steeper on the descending side), reduces maximum range and height, and lowers the optimal launch angle below 45°. For fast objects (bullets, baseballs, cannon shells), air resistance is significant and numerical methods are used to integrate the trajectory. For slow, heavy objects over short distances, the ideal model is fine.
What is the difference between range and displacement? ▾
Range is the horizontal distance from launch to landing point, measured along the ground. Total displacement is the straight-line distance from launch to landing position (which may differ if launched from a height). For ground-level launches, the vertical displacement is zero at landing, so horizontal range = horizontal displacement. From a cliff, the total displacement = √(range² + h²) at an angle below horizontal.
How do I find the velocity at any point during flight? ▾
At time t: vx = v₀·cosθ (constant) and vy = v₀·sinθ − g·t (decreasing linearly). Total speed = √(vx² + vy²) and direction angle below horizontal = arctan(|vy|/vx) if descending. At launch, speed = v₀ at angle θ. At peak, speed = vx = v₀·cosθ. At landing (same height), speed = v₀ (by energy conservation), angle θ below horizontal.

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