Projectile Time of Flight
Also known as hang time
Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!
Learning zone
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.
- = Time of flight
- = Launch speed
- = Launch angle
- 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