Projectile Motion Calculator
Calculate the trajectory, range, maximum height, and time of flight for projectile motion problems.
Initial Conditions
Velocity Components:
Horizontal Range
40.77 m
Maximum height: 10.19 m
Trajectory Points:
Impact Details
Projectile Motion Equations
Position
x(t) = v₀ₓ × t = v₀ × cos(θ) × t
y(t) = h₀ + v₀ᵧ × t - ½gt²
Velocity
vₓ = v₀ₓ (constant)
vᵧ = v₀ᵧ - g × t
Maximum Height
H = h₀ + v₀ᵧ² / (2g)
Range (from ground)
R = v₀² × sin(2θ) / g
Key Facts
- • Maximum range is achieved at 45° launch angle (from ground level)
- • Complementary angles (e.g., 30° and 60°) give the same range
- • Horizontal velocity remains constant (no air resistance)
- • Time to reach maximum height = v₀ᵧ / g
- • Total flight time is symmetric when launching from and landing at the same height
What is Projectile Motion?
Projectile motion describes the path of an object that is launched into the air and moves under the influence of gravity alone (ignoring air resistance). The key insight is that horizontal and vertical motions are independent.
| Component | Acceleration | Velocity | Behavior |
|---|---|---|---|
| Horizontal (x) | aₓ = 0 | vₓ = constant | Uniform motion |
| Vertical (y) | aᵧ = -g = -9.81 m/s² | vᵧ changes | Free fall |
The resulting path is a parabola. This applies to thrown balls, launched rockets (initially), cannonballs, and jumping animals.
Initial Velocity Components
Where:
- v₀= Initial velocity (m/s)
- θ= Launch angle from horizontal
- v₀ₓ= Horizontal velocity component
- v₀ᵧ= Vertical velocity component
Projectile Motion Equations
The complete equations for projectile motion (launched from origin):
| Quantity | Horizontal (x) | Vertical (y) |
|---|---|---|
| Position | x = v₀ₓt = v₀cos(θ)t | y = v₀ᵧt - ½gt² = v₀sin(θ)t - ½gt² |
| Velocity | vₓ = v₀ₓ = v₀cos(θ) | vᵧ = v₀ᵧ - gt = v₀sin(θ) - gt |
| Acceleration | aₓ = 0 | aᵧ = -g |
At any time: Speed |v| = √(vₓ² + vᵧ²), Direction θ = tan⁻¹(vᵧ/vₓ)
Position Equations
Where:
- x(t)= Horizontal position at time t
- y(t)= Vertical position at time t
- t= Time since launch (s)
- g= 9.81 m/s² (acceleration due to gravity)
Key Values: Range, Height, and Time
Important derived quantities for projectile motion:
| Quantity | Formula | Condition |
|---|---|---|
| Time of flight | T = 2v₀sin(θ)/g | Returns to launch height |
| Maximum height | H = v₀²sin²(θ)/(2g) | When vᵧ = 0 |
| Horizontal range | R = v₀²sin(2θ)/g | Level ground |
| Time to max height | t = v₀sin(θ)/g | Half of total flight time |
| Launch Angle | sin(2θ) | Range (relative) | Height (relative) |
|---|---|---|---|
| 15° (or 75°) | 0.50 | 50% | 7% / 93% |
| 30° (or 60°) | 0.87 | 87% | 25% / 75% |
| 45° | 1.00 | 100% (maximum) | 50% |
Complementary angles (like 30° and 60°) give the same range but different heights and flight times.
Range and Maximum Height
Where:
- R= Horizontal range (m)
- H= Maximum height (m)
- T= Total time of flight (s)
Optimal Launch Angle
The optimal angle depends on what you're trying to achieve:
| Goal | Optimal Angle | Why |
|---|---|---|
| Maximum range (level) | 45° | sin(2×45°) = sin(90°) = 1 |
| Maximum height | 90° | All velocity is vertical |
| Fastest arrival | Lower angles | More horizontal velocity |
| Target above launch | > 45° | Need more vertical component |
| Target below launch | < 45° | Gravity assists range |
With air resistance: The optimal angle for maximum range drops to about 35-40° because air drag reduces horizontal velocity over time.
Trajectory Equation
The trajectory equation gives y as a function of x (eliminating time):
| Form | Equation | Use |
|---|---|---|
| General trajectory | y = x·tan(θ) - gx²/(2v₀²cos²(θ)) | Path shape |
| Simplified | y = x·tan(θ) - (g/2v₀ₓ²)x² | Using v₀ₓ directly |
| Vertex form | y = H - (g/2v₀ₓ²)(x - R/2)² | Shows max height at midpoint |
This is a parabola opening downward. The shape depends only on v₀ and θ (assuming constant g).
Trajectory Equation
Where:
- y= Vertical position (m)
- x= Horizontal position (m)
- θ= Launch angle
- v₀= Initial speed (m/s)
Special Cases
Common projectile motion scenarios:
| Case | Launch Angle | Key Equations |
|---|---|---|
| Horizontal launch | θ = 0° | x = v₀t, y = -½gt², R = v₀√(2h/g) |
| Vertical launch | θ = 90° | R = 0, H = v₀²/(2g), T = 2v₀/g |
| Dropped object | v₀ = 0 | y = -½gt², t = √(2h/g) |
| Launched from height h | Any | Solve y = h + v₀sin(θ)t - ½gt² = 0 |
| Situation | Time of Flight | Range |
|---|---|---|
| Drop from height h | t = √(2h/g) | 0 |
| Horizontal throw from h | t = √(2h/g) | R = v₀√(2h/g) |
| Throw up from h | Longer (solve quadratic) | Depends on angle |
Real-World Applications
Projectile motion principles apply to many activities:
| Application | Typical Parameters | Considerations |
|---|---|---|
| Basketball free throw | v₀ ≈ 7 m/s, θ ≈ 50° | Optimal angle higher due to hoop height |
| Long jump | v₀ ≈ 10 m/s, θ ≈ 20° | Optimal lower than 45° due to technique |
| Golf drive | v₀ ≈ 70 m/s, θ ≈ 10-15° | Backspin creates lift (not simple projectile) |
| Artillery | v₀ ≈ 800 m/s | Air resistance, Earth's rotation matter |
| Javelin throw | v₀ ≈ 30 m/s, θ ≈ 35° | Aerodynamic design |
| Water fountain | Various | Decorative parabolic arcs |
Note: In real applications, air resistance, spin, and other factors can significantly affect trajectories.
Worked Examples
Basic Projectile Problem
Problem:
A ball is launched at 20 m/s at a 45° angle. Find the range, maximum height, and time of flight.
Solution Steps:
- 1Identify values: v₀ = 20 m/s, θ = 45°, g = 9.81 m/s²
- 2Range: R = v₀²sin(2θ)/g = 20²×sin(90°)/9.81 = 400/9.81 = 40.8 m
- 3Max height: H = v₀²sin²(θ)/(2g) = 400×0.5/(2×9.81) = 10.2 m
- 4Time of flight: T = 2v₀sin(θ)/g = 2×20×0.707/9.81 = 2.88 s
Result:
Range = 40.8 m, Max height = 10.2 m, Time = 2.88 s
Horizontal Launch from Height
Problem:
A ball is thrown horizontally at 15 m/s from a 45 m tall building. How far from the base does it land?
Solution Steps:
- 1This is horizontal launch: θ = 0°, v₀ = 15 m/s, h = 45 m
- 2Time to fall: t = √(2h/g) = √(2×45/9.81) = √9.17 = 3.03 s
- 3Horizontal distance: x = v₀t = 15 × 3.03 = 45.4 m
- 4Final vertical velocity: vᵧ = gt = 9.81 × 3.03 = 29.7 m/s
Result:
Lands 45.4 m from base, vertical impact speed = 29.7 m/s
Finding Launch Angle
Problem:
A golf ball must travel 150 m. If initial speed is 50 m/s, what launch angle is needed?
Solution Steps:
- 1Use range formula: R = v₀²sin(2θ)/g
- 2Rearrange: sin(2θ) = Rg/v₀² = 150×9.81/2500 = 0.589
- 32θ = sin⁻¹(0.589) = 36.1° or 143.9°
- 4θ = 18° or 72° (complementary angles give same range)
- 518° gives flatter, faster trajectory; 72° gives higher arc
Result:
Launch angle = 18° or 72° (same range, different paths)
Tips & Best Practices
- ✓Split motion into x (constant velocity) and y (constant acceleration) components
- ✓Maximum range on level ground is at 45° launch angle
- ✓Time of flight depends only on vertical motion: T = 2v₀sin(θ)/g
- ✓Complementary angles (like 30° and 60°) give the same range but different heights
- ✓At maximum height, vertical velocity is zero but horizontal velocity is unchanged
- ✓For horizontal launches, time in air depends only on height: t = √(2h/g)
- ✓The trajectory is a parabola—every projectile follows this shape (ignoring air resistance)
Frequently Asked Questions
Sources & References
Last updated: 2026-01-22