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Up at v0, then back. H_max = v0²/(2g) — enough for the classic homework problem.
Part 4 of Physics without mysteries. Comparison hub: throws side by side.
An upward vertical throw starts with v0 opposite g. At the apex v = 0, so from v² = v02 − 2gH you get H_max = v02/(2g). Rise time is t_up = v0/g; the full up-and-back flight (to the same height) is 2 t_up.
Energetically: at launch Ek = ½mv02; at the top the same amount sits in Ep = mgH. Free fall is the v0 = 0 limit — no upward H_max, only downward distance.
Apollo 15 (1971) dropped a hammer and a feather on the Moon to show that without air both fall together. School problems still use constant g and zero drag.
| Goal | Formula | Note |
|---|---|---|
| H_max (upward) | H = v₀²/(2g) | v = 0 at the apex |
| Rise time | t = v₀/g | Same again on the way down |
| State at time t | v = v₀ − gt, h = v₀t − ½gt² | Axis upward |
| Fall with v₀ = 0 | h = ½gt² | Free-fall / free-fall-time calculator |
You throw a ball straight up at v0 = 10 m/s. Take g = 9.81 m/s². Find H_max.
Steps
Answer: H_max ≈ 5.10 m
