Article
Leave an edge with speed only in x. Flight time comes from the fall; range from constant vx.
Part 5 of Physics without mysteries. Comparison: throw hub.
In a horizontal throw v0 has only an x component. Vertically you start at zero speed and fall a height h: t = √(2h/g). Horizontally you cruise at constant v0, so Z = v0·t.
It is the angled-throw limit as α → 0° from a height. If someone gives an angle and h0, use full projectile motion instead of forcing the simplification.
Bridge example: h = 20 m, v0 = 10 m/s → t ≈ 2.02 s and Z ≈ 20.2 m (g = 9.81). School problems ignore air drag.
| Quantity | Formula | Source |
|---|---|---|
| Flight time | t = √(2h/g) | Vertical fall with vᵧ₀ = 0 |
| Range | Z = v₀·t | Constant vₓ = v₀ |
| Path | y = h − ½gt², x = v₀t | Parabola in the plane |
| α = 0, h₀ > 0 | full projectile model | When you also want landing vᵧ |
From a bridge of height h = 20 m you throw horizontally at v0 = 10 m/s. Take g = 9.81 m/s². Find flight time and range.
Steps
Answer: t ≈ 2.02 s, Z ≈ 20.2 m
