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Motion in Two Dimensions: Projectiles

The secret of a thrown ball's arc is that its horizontal and vertical motions are completely independent — and once you see that, you can predict where it lands.

The one big idea: x and y are independent

Two-dimensional motion looks harder than one-dimensional motion, but there is a single idea that tames it: the horizontal and vertical motions proceed independently, each obeying its own one-dimensional equations, tied together only by sharing the same clock. Break any 2D velocity into perpendicular components and you can treat each direction on its own.

Setting up a projectile

Projectile motion is the motion of an object launched into the air and then moving under gravity alone (we idealize away the air). Launch it with speed v_0 at angle \theta above the horizontal, and split that launch velocity into a horizontal part v_{0x} = v_0\cos\theta and a vertical part v_{0y} = v_0\sin\theta.

Along the arc the velocity vector tips over: its horizontal part stays constant while its vertical part shrinks, reverses at the peak, and grows on the way down. Only vy changes — gravity acts purely vertically.

The projectile equations

Now apply what we already know to each direction. Horizontally there is no acceleration, so the motion is uniform. Vertically the acceleration is -g, so it is free fall. Written out with the launch at the origin:

x = (v_0\cos\theta)\,t, \qquad y = (v_0\sin\theta)\,t - \tfrac{1}{2}\,g\,t^2

Horizontal: constant velocity. Vertical: free fall. The shared variable is the time t — that is the only thread linking the two.

Eliminating t between these two shows that y is a quadratic in x: the path is a parabola. That graceful arc of a thrown ball or a fountain's water is not a coincidence — it is what 'constant horizontal speed plus constant vertical acceleration' always draws. The shape of the path is the trajectory.

Range, height, and time of flight

For a projectile landing back at its launch height, a little algebra on those two equations gives the three quantities people usually care about. The range R (horizontal distance travelled), the maximum height H, and the total time of flight T:

R = \frac{v_0^2 \sin 2\theta}{g}, \qquad H = \frac{v_0^2 \sin^2\theta}{2g}, \qquad T = \frac{2\,v_0 \sin\theta}{g}

Range, maximum height, and time of flight for launch and landing at the same height. Range is largest at θ = 45°, where sin 2θ = 1.

Change the launch speed and angle and watch the parabola, range, peak and flight time respond. Notice that complementary angles — say 30° and 60° — give the same range, and that 45° reaches farthest.