Projectile Time of Flight
Also known as hang time
Worked example: 25 m/s at 30° → 2.549 s — press Try an example to run it live, then adjust anything.
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The height of a throw →
Grade 11Grade 11 Math — Functions & Applications
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Projectile Time of Flight explained
Hang time depends only on the vertical launch velocity: the projectile takes v₀ sin θ ⁄ g to reach the top and exactly as long to fall back, giving T = 2v₀ sin θ ⁄ g. Launch at 25 m/s and 30° and the vertical component is 12.5 m/s, so the flight lasts 2 × 12.5 ⁄ 9.80665 ≈ 2.55 s. Nothing about the horizontal speed appears — which is Galileo's independence principle in its most testable form, the same insight behind his claim that a stone dropped from a moving ship's mast lands at the mast's foot.
Two cautions. First, this assumes the landing height equals the launch height; throw off a cliff or off a 2 m-tall shoulder and the real flight is longer, requiring the full quadratic. Second, the formula is symmetric, so a given time of flight matches two speeds only through the angle — solving for θ returns the principal arcsin branch (0° to 90°), and the supplementary angle 180° − θ gives the identical hang time on a lower, faster trajectory.
Projectile Time of Flight formula
- = Time of flight (s)
- = Launch speed (m/s)
- = Launch angle (°)
Missing one of these? Work it out first, then come back
- Time of flight — Speed, Distance & Time, Final Velocity (Uniform Acceleration)
- Launch speed — Projectile Range on Level Ground, Projectile Maximum Height
- Launch angle — Projectile Range on Level Ground, Projectile Maximum Height