How to use this calculator
Enter the launch speed and angle; gravity is pre-filled with standard gravity (9.80665 m/s²) and can be changed for the Moon (1.62 m/s²) or Mars (3.71 m/s²). The calculator splits the launch velocity into horizontal and vertical parts and returns the time of flight, maximum height and range.
Any other combination of two or three quantities works too. Enter a target range and a launch speed to find the angle (the lower of the two angles that reach that distance is shown — the other is 90° minus it), or a range and peak height to recover both speed and angle from a video of a throw.
The formula
- v₀
- launch speed
- θ
- launch angle
- vₓ
- horizontal velocity
- v_y
- initial vertical velocity
- T
- time of flight
- H
- maximum height
- R
- horizontal range
- g
- gravitational acceleration
Worked example
A ball is kicked at 20 m/s, 30° above level ground. How far does it go, how high does it rise, and how long is it in the air?
- Split the velocity: vₓ = 20 cos 30° = 17.32 m/s, v_y = 20 sin 30° = 10 m/s.
- Time of flight T = 2v_y / g = 2 × 10 / 9.807 = 2.04 s; peak height H = v_y² / 2g = 5.10 m.
- Range R = vₓ × T = 17.32 × 2.039 = 35.3 m.
The ball lands about 35 m away after 2 seconds, peaking at just over 5 m.
Assumptions and limits
- Launch and landing are at the same height (level ground). A raised launch point increases range and flight time.
- Air resistance is neglected, which is accurate for dense, slow objects but overestimates the range of light or fast ones (a baseball travels well under the vacuum figure).
- Gravity is uniform over the trajectory.
Frequently asked questions
What angle gives the maximum range?
45° on level ground without air resistance, because sin 2θ peaks at θ = 45°. With drag, the optimum is lower — often 35–40° for sports balls.
Why do two angles give the same range?
sin 2θ = sin(180° − 2θ), so θ and 90° − θ produce the same range. The steeper launch goes higher and stays in the air longer.
How is maximum height related to time of flight?
On level ground H = gT²/8, so a 2-second hang time means the object rose about 4.9 m.
Sources
- OpenStax University Physics Volume 1, §4.3 Projectile Motion
- HyperPhysics (Georgia State University) — Trajectories
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