Formula & Calculator
Projectile Maximum Height
Calculates the maximum height a projectile reaches during its flight based on its launch speed and angle.
Interpretation
Maximum height of a projectile: H = v²·sin²(θ)/(2g). It is the vertical distance reached. Example: v=20 m/s, θ=45° → H = 400×0.5/(2×9.81) ≈ 10.2 m.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| H | Maximum height reached | m |
| v | Initial launch speed | m/s |
| theta | Launch angle above horizontal | degrees |
| g | Gravitational acceleration | 9.81 m/s2 |
What it means
The maximum height of a projectile is the highest vertical point reached during its flight. It occurs when the vertical velocity becomes zero. The formula H = v² sin²θ / (2g) is derived from the kinematic equation v_y² = u_y² + 2a(y‑y₀). This height depends on the initial speed and launch angle. It is used to determine the clearance needed over obstacles, to design parabolic trajectories, and in sports to evaluate performance. In engineering, it helps in designing projectile paths and in safety assessments. Understanding this formula is essential for analysing vertical motion components.
Worked example
Projectile Maximum Height – Two Examples
Real‑World| Parameter | Value |
|---|---|
| v | 20 m/s |
| θ | 45° |
| Parameter | Value |
|---|---|
| v | 30 m/s |
| θ | 30° |
Common mistakes
- Launch angle θ: With respect to horizontal.
- Initial speed v: Magnitude – not vertical component.
- Maximum height H: Vertical displacement from launch point – only valid for launch from ground level.
- Units: v in m/s, g in m/s² → H in m.
- Air resistance: Ignored; real maximum height is lower.
Applications
The maximum height of a projectile, H = v² sin²θ / (2g), is the highest point reached during its flight. This formula is used to assess the clearance of obstacles, to design water fountains, and to analyse the performance of fireworks. In sports, it helps evaluate the arc of a basketball shot or a high jump. In safety engineering, it determines the maximum height a falling object could reach if projected upward. The formula also aids in the design of particle accelerators and in the analysis of ballistic impacts. By understanding projectile height, engineers can ensure that trajectories remain within safe limits and that structures are not hit by airborne objects.
- Obstacle clearance analysis in projectile motion
- Design of water fountains and decorative jets
- Sports technique optimisation (shot put, basketball)
- Safety assessments for airborne debris
- Ballistic impact and penetration studies
Frequently Asked Questions
The maximum height reached by a projectile launched with initial speed v at angle θ is H = v²·sin²θ / (2g). This is the vertical distance above the launch point.
Using the full launch speed v instead of only its vertical component (v·sinθ). The horizontal component does not affect the maximum height.
The height is maximized when θ = 90° (vertical launch). Then H = v²/(2g). For any other angle, the height is lower.
Height is proportional to v². Doubling the speed quadruples the maximum height.
The time to reach maximum height is t_peak = v·sinθ / g. The maximum height is the displacement during this time: H = ½g·t_peak² (since initial vertical velocity is v·sinθ and final is zero).
Vertical component = 20·sin30° = 10 m/s. H = (10)² / (2×9.81) ≈ 5.10 m.
Yes, air resistance reduces the maximum height because it opposes the motion, reducing the vertical speed.
The formula still gives the height above the launch point. To get the height above the ground, add the launch height (if known).