How does projectile motion work?

How does projectile motion work?

A projectile is any object moving through the air with gravity as the only force acting on it. Its motion splits into two parts that do not affect each other: the horizontal part travels at a constant speed, and the vertical part accelerates downward at $g \approx 9.81\ \text{m/s}^2$. Add those two motions together and the object traces a parabola.

Everything else — how far it lands, how high it climbs, how long it stays up — follows from those two rules plus the launch speed $v_0$ and angle $\theta$.

Why are the two motions independent?

Gravity pulls straight down, so it can only change the vertical velocity. It has no horizontal component, which means the horizontal velocity a projectile is launched with is the horizontal velocity it keeps for the whole flight (ignoring air). This is why a bullet fired horizontally and a bullet dropped from the same height hit the ground at the same moment: their vertical motions are identical, and the horizontal motion of the fired bullet does not delay the fall.

The equations of projectile motion

Resolve the launch velocity into components — $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$ — then apply constant-velocity motion horizontally and constant-acceleration motion vertically:

Eliminating $t$ between those two equations gives $y$ as a function of $x$ — and the result is a quadratic in $x$, which is the equation of a parabola. That is why the path is parabolic rather than, say, circular.

Range, maximum height and time of flight

Three numbers describe the whole arc. Setting $y = 0$ in the vertical equation and solving the quadratic gives the time of flight; the horizontal equation then gives the range; and the apex is where the vertical velocity $v_y = v_0\sin\theta - g t$ passes through zero.

The range formula contains $\sin 2\theta$, which is largest when $2\theta = 90°$, i.e. $\theta = 45°$. That is the famous result — but it only holds when the projectile lands at its launch height and there is no drag. Launch from a cliff and the best angle drops below $45°$; add air resistance and it drops further still.

What changes with drag and wind?

The clean parabola assumes gravity is the only force. Turn on drag in the simulation and the air pushes back opposite to the velocity, growing with speed — the trajectory becomes lopsided, the descent steeper than the climb, and the range shorter than the ideal formula predicts. Wind adds a constant horizontal push that stretches the range downwind and compresses it upwind. With either enabled, the predicted landing point in the readout starts to diverge from where the ball actually lands, which is the whole point of the comparison.

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