A projectile is an object whose motion, after launch, is modeled using the approximately constant downward gravitational acceleration near Earth's surface while air resistance and other effects are neglected.

The central idea is that horizontal and vertical motion share the same time but follow independent component equations.

Choose the $x$-axis horizontal and the $y$-axis upward. If a projectile is launched with speed $v_0$ at angle $\theta$ above the horizontal, its initial velocity components are

$$v_{0x}=v_0\cos\theta,$$

$$v_{0y}=v_0\sin\theta.$$

Under the uniform-gravity model,

$$a_x=0,\qquad a_y=-g.$$

Therefore the velocity components are

$$v_x=v_0\cos\theta,$$

$$v_y=v_0\sin\theta-gt,$$

and the position is

$$x=x_0+v_0\cos\theta,t,$$

$$y=y_0+v_0\sin\theta,t-\frac12gt^2.$$

The horizontal velocity remains constant while gravity continuously changes the vertical velocity.

The trajectory is parabolic

When $\cos\theta\neq0$, solve the horizontal equation for time:

$$t=\frac{x-x_0}{v_0\cos\theta}.$$

Substituting into the vertical equation gives

$$y-y_0 =(x-x_0)\tan\theta -\frac{g(x-x_0)^2}{2v_0^2\cos^2\theta}.$$

This is a quadratic function of $x$, so the ideal projectile trajectory is a parabola.

Launch and landing at the same height

Suppose the projectile returns to its launch height, so $y=y_0$. Besides the launch time $t=0$, the vertical equation gives the flight time

$$t_f=\frac{2v_0\sin\theta}{g}.$$

The maximum height above the launch point occurs when the vertical velocity reaches zero:

$$0=v_0\sin\theta-gt_{\rm top},$$

so

$$t_{\rm top}=\frac{v_0\sin\theta}{g}.$$

Substitution gives

$$h_{\max}=\frac{v_0^2\sin^2\theta}{2g}.$$

The horizontal range is

$$R=(v_0\cos\theta)t_f =\frac{v_0^2\sin(2\theta)}{g}.$$

For a fixed launch speed, equal launch and landing heights, and negligible drag, the range is largest when

$$\sin(2\theta)=1,$$

which gives

$$\theta=45^\circ.$$

The $45^\circ$ result is therefore conditional, not a universal rule for every projectile problem.

Worked example

A ball is launched from ground level at

$$v_0=20,\mathrm{m/s}$$

and

$$\theta=30^\circ.$$

Its flight time is

$$t_f=\frac{2(20)\sin30^\circ}{9.81}\approx2.04,\mathrm s.$$

Its maximum height is

$$h_{\max}=\frac{(20)^2\sin^2 30^\circ}{2(9.81)}\approx5.10,\mathrm m,$$

and its range is

$$R=\frac{(20)^2\sin60^\circ}{9.81}\approx35.3,\mathrm m.$$

The horizontal and vertical calculations are not separate motions occurring at different times. They are two components of the same trajectory, linked by the same value of $t$.

Demonstration: testing the range prediction

Walter Lewin tests the predicted projectile range experimentally, including the uncertainty of the launch angle and speed, rather than treating the $45^\circ$ result as algebra alone.