# Launch Angle

Wave · y = f(x, t)

`y = max(x · tan(deg(45) + deg(30) · sin(t)) −((9.8x²)/(2 · v₀² · cos(deg(45) + deg(30) · sin(t))²)), 0)`

[Open in the app](https://www.wavelace.com/app#p=131) · [This page](https://www.wavelace.com/presets/launch-angle)

### What it draws

This is the path of a projectile thrown from the origin, with `x` the distance downrange and the value the height. Eliminating the time of flight from the two equations of motion leaves the height as `x·tan(θ) − g·x²/(2·v₀²·cos²(θ))`, a straight rise at the launch angle with a parabolic droop subtracted from it. Gravity is written into the formula as the number 9.8, the value on the Earth, so the one letter left is `v₀`, the launch speed, opening at 8.

The launch angle is not a letter but the clock. `θ = 45° + 30°·sin t` swings between 15° and 75° and back once every `2π ≈ 6.28` seconds at `Speed` 1. The outer `max(…, 0)` is the ground. Past the landing point the parabola dives steeply, so the floor cuts it off and each frame reads as flat ground, one arc, flat ground again.

### Why 45° wins

The parabola returns to zero at `x = v₀²·sin(2θ)/g`, which is 6.53 at 45° and the longest throw the sweep reaches. Because `sin(2θ)` is unchanged by swapping `θ` for `90° − θ`, the shallow and the steep throw land together: 15° and 75° both reach 3.27. The apex is `v₀²·sin²(θ)/(2g)`, which is 0.22 for the flat throw and 3.05 for the lobbed one, a factor of 14 between two arcs of identical range.

That symmetry shows in the timing: the angle passes 45° twice in each swing, at `t = 0` and `t = π`, so the landing point runs out and back twice while the arc rises and falls once.

### What the launch speed does

The letter changes the size of the picture and not the shapes in it, because the range depends on speed and gravity only through `v₀²/g`. Halving `v₀` shrinks every arc of the sweep to a quarter, and doubling it stretches them fourfold. Only the square of the speed appears, so the sign is immaterial and `−8` throws exactly as `8` does. At `v₀ = 0` the droop is infinite everywhere, the zero floor catches all of it, and only flat ground is left.

### Try

- Drag `v₀` to 10: the 45° throw reaches 10.20, because range follows the square of the launch speed, and still lands inside the opening view.
- Drag `v₀` to 4: every arc shrinks to a quarter of the one beside it, the longest throw reaching 1.63.
- Change the `9.8` to `3.7`, roughly the gravity of Mars: the same throw now carries 17.30, so raise `Span of x` to 18 to watch it land.
- Change both `30°` to `44°`: the sweep then runs from 1° to 89°, and both ends of it barely clear the launch point.
- `Top` in the deck: from above, the ribbon's outer edge traces the landing point against time, touching the near mark twice a swing as the shallow throw and the steep one land together.

### Read more

- [Projectile motion](https://en.wikipedia.org/wiki/Projectile_motion)
- [Range of a projectile](https://en.wikipedia.org/wiki/Range_of_a_projectile)
- [Trajectory](https://en.wikipedia.org/wiki/Trajectory)
- [Parabola](https://en.wikipedia.org/wiki/Parabola)