Projectile Max Height

Maximum height: h_max = (v₀·sin θ)² / (2g)

Launch a projectile with initial speed v₀ at angle θ above the horizontal and ignore air resistance, and the highest point it reaches above the launch level works out to h_max = (v₀·sin θ)² / (2g). It gets there at time t = v₀·sin θ / g. Where does this come from? Plain kinematics. At the top of the arc the vertical velocity has dropped to zero, so v_y² = (v₀·sin θ)² − 2g·h_max = 0, and solving for h_max gives the formula. Notice that in a vacuum the answer is independent of mass, since gravity pulls on every body at the same rate. For a fixed speed you reach the absolute ceiling at θ = 90°, throwing straight up. To put numbers on it: v₀ = 50 m/s at θ = 60° gives h_max ≈ 95.7 m at t ≈ 4.42 s, while the same speed fired vertically tops out at 127.4 m. Mass starts to count only once you add air drag, where the ratio of mass to cross-sectional area (the ballistic coefficient) decides how much speed bleeds away on the climb.

Applications

Fireworks lean on it for safe altitude and burst timing. So does pole vault (Mondo Duplantis's 6.23 m world record), platform diving (a 10 m pool platform buys about 1.4 s of free fall), scoring in trampoline and gymnastics, model rocketry, and the physics labs where most people first meet the equation.

FAQ

Why does mass not appear in the formula? In a vacuum every body falls with the same gravitational acceleration (Galileo's principle), so mass simply cancels out of the equation of motion.

Which angle gives the highest h_max? A vertical launch, θ = 90°, since sin(90°) = 1. Tilt the launch at all and the vertical component v₀·sin θ shrinks.

When does mass matter in practice? Once air drag enters the scene. A heavier object of the same shape has a higher ballistic coefficient, so it loses less speed on the way up and climbs higher than a lighter twin would.