Playground
Animated projectile arc with adjustable angle and speed. Shows parabolic trajectory and range marker.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Rangeoutput Horizontal distance traveled by the projectile | m | L | 0 – 5000 | |
| Launch Speed Initial speed of the projectile | m/s | LT⁻¹ | 1 – 200 | |
| Launch Angle Angle of launch above horizontal (in degrees) | ° | dimensionless | 0 – 90 | |
| Gravitational Acceleration Acceleration due to gravity at Earth's surface | m/s² | LT⁻² | 9.8 – 9.8 |
Deep dive
Derivation
Split initial velocity into vₓ = v·cosθ and vᵧ = v·sinθ. Time of flight: T = 2v·sinθ/g. Range: R = vₓ·T = v·cosθ · 2v·sinθ/g = v²·sin(2θ)/g.
Experimental verification
Ballistic range tables, artillery firing data, and high-speed camera trajectory tracking.
Common misconceptions
- Air resistance makes optimal angle less than 45° in practice
- Complementary angles (30° and 60°) give equal range but different flight times
- The formula assumes flat ground — uphill/downhill changes the optimal angle
Real-world applications
- Sports trajectory optimization
- Artillery and ballistics
- Sprinkler coverage design
- Long jump technique
Worked examples
Optimal soccer kick
Given:
- v:
- 25
- theta:
- 45
- g:
- 9.8
Find: R
Solution
R = v²sin(2×45°)/g = 625×1/9.8 = 63.8 m
Complementary angles give equal range
Given:
- v:
- 20
- theta_1:
- 30
- theta_2:
- 60
- g:
- 9.8
Find: R for both angles
Solution
Both give R = 400×sin(60°)/9.8 = 400×0.866/9.8 = 35.3 m
Scenarios
What if…
- scenario:
- What if on the Moon (g = 1.62)?
- answer:
- Range increases by factor 9.8/1.62 = 6.05×. A 63.8 m kick becomes 386 m on the Moon.
- scenario:
- What if air resistance is included?
- answer:
- Optimal angle drops below 45° (typically 30°–40° depending on drag). Range decreases significantly at high speeds.
- scenario:
- What if launched from a cliff (height h)?
- answer:
- The formula changes: R = (v²/2g)[sin(2θ) + √(sin²(2θ) + 8gh/v²·cos²θ)]. Higher launch point always increases range.
Limiting cases
- condition:
- θ = 0° or 90°
- result:
- R = 0
- explanation:
- Horizontal launch has no hang time; vertical launch has no horizontal component.
- condition:
- θ = 45°
- result:
- R = v²/g (maximum)
- explanation:
- sin(90°) = 1 gives the maximum range.
- condition:
- g → 0
- result:
- R → ∞
- explanation:
- Without gravity, the projectile travels in a straight line forever.
Context
Galileo Galilei · 1638
Derived in Discorsi e dimostrazioni matematiche, where Galileo analyzed parabolic trajectories of cannonballs.
Hook
A soccer player kicks at 25 m/s. At what angle does the ball travel the farthest?
R = v²sin(2θ)/g is maximized when sin(2θ) = 1, i.e., θ = 45°. At 25 m/s: R = 625/9.8 ≈ 63.8 m.
Dimensions:
- lhs:
- R → [L]
- rhs:
- v²·sin(2θ)/g → [LT⁻¹]²/[LT⁻²] = [L²T⁻²]/[LT⁻²] = [L]
- check:
- Both sides are [L] = meters. sin(2θ) is dimensionless. ✓
Validity: Valid for flat ground, no air resistance, uniform g = 9.8 m/s², and non-relativistic launch speeds.