Case 1
- Launch speed
- 20m/s
- Launch angle
- 45°
10.197m
Open with these valuesProjectile Maximum Height Calculator
10.197m
Result: 10.197 mOnly the vertical share of the launch speed lifts a projectile, so the peak sits at v² × sin²θ ÷ (2g). A 20 m/s throw at 45° tops out at 10.197 m; the same speed straight up reaches exactly twice that. Air resistance is ignored, so read the number as a clean upper bound.
Worked examples
Case 2
- Launch speed
- 10m/s
- Launch angle
- 30°
1.275m
Open with these valuesCase 3
- Launch speed
- 50m/s
- Launch angle
- 60°
95.598m
Open with these valuesHow it's calculated
h = v² × sin²θ ÷ (2 × g)
- StepEnter the launch speed in metres per second — divide km/h by 3.6.
- StepEnter the angle above the horizontal, between 0° and 90°.
- ResultRead the peak height in metres above the launch point.
Reference table
| Speed, angle | Launch | Peak height (m) |
|---|---|---|
| 20, 0 | Flat, nothing rises | 0.000 |
| 10, 30 | Shallow | 1.275 |
| 15, 37 | Shallow | 4.155 |
| 20, 45 | Best for range | 10.197 |
| 20, 90 | Straight up, the ceiling | 20.394 |
| 50, 60 | Fast and steep | 95.598 |
Questions
How do I calculate the maximum height of a projectile?
Use h = (v² × sin²θ) / (2g): square the launch speed, multiply by the square of the sine of the angle, then divide by twice gravity. With v in m/s and θ in degrees the height comes out in metres. A 20 m/s launch at 45° reaches (20² × sin²45°) / (2 × 9.80665) ≈ 10.197 m.
What is the maximum height of a projectile?
It is the highest point a launched object reaches before gravity brings it back down — the very top of its arc, measured above the launch point. Only the vertical part of the launch speed contributes, so a steeper angle gives a higher peak for the same speed.
Which launch angle gives the greatest height?
A launch straight up, at 90°, gives the greatest possible height for a given speed, because sin²90° = 1 puts all the velocity into rising. As you flatten the angle the height falls with sin²θ, reaching half the maximum at 45° and zero at a horizontal 0° launch.
Why don't maximum height and range peak at the same angle?
Height depends on the vertical speed (sin θ) and keeps growing up to 90°, while horizontal range depends on both components and peaks at 45°. The steeper the angle, the more speed goes into rising rather than travelling forward.
Does this account for air resistance?
No. This calculator uses the idealised projectile-motion formula with no air resistance and a launch from ground level, so it gives the height above the release point. Real projectiles lose energy to drag and rise a little less, with the gap growing at higher speeds and larger surface areas.
Sources and last check
Information, not professional advice.