Motion in Two or Three Dimensions

Roll a ball off a table and, at the same instant, drop a second ball from the same height. The rolled ball lands farther away, yet the two hit the floor together. Its sideways motion did nothing to its fall, and that independence is what makes motion in a plane solvable: you treat it as two one-dimensional problems that share a clock.

Split the motion into components

Write the position as $\vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j}$ (add $z(t)\,\hat{k}$ in three dimensions). Differentiating a vector differentiates each component, so

Nothing in the $x$ equations involves $y$, and nothing in the $y$ equations involves $x$. So every tool from topics 1.2 and 1.3 (derivatives, integrals, graphs, and the constant-acceleration equations when they apply) works on each axis by itself. The only thing the two axes share is the time $t$, which is how you pass information between them: find the time from one axis, use it in the other.

The two components need not behave alike. A toy boat crossing a pool might have a constant $v_x = 3.0\ \text{m/s}$ while a motor gives it $v_y = (1.2\ \text{m/s}^3)\,t^2$. Then $a_x = 0$ and $a_y = (2.4\ \text{m/s}^3)\,t$: one component has no acceleration and the other has an acceleration that changes with time. Each is still a 1D problem; the second just needs calculus rather than the constant-acceleration equations.

When the components are found, recombine them as in topic 1.1: speed $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$ and direction $\tan^{-1}\left(\frac{v_y}{v_x}\right)$, with the quadrant read from the signs.

Perpendicular motions are independent

Changing the motion along one axis leaves the perpendicular motion alone. A puck sliding east on frictionless ice at $3.0\ \text{m/s}$ that gets a quick kick due north picks up a north component of velocity, but its east component stays $3.0\ \text{m/s}$, because nothing acted along the east-west line. In Unit 2 this becomes a statement about forces: a force along $y$ produces acceleration only along $y$.

The strobe picture below shows the opening experiment. Every $0.10\ \text{s}$, the launched ball and the dropped ball are at the same height, because both start with $v_y = 0$ and have the same $a_y = -g$. The launched ball's horizontal spacing is even, since nothing changes its $v_x$.

Projectile motion

A projectile is an object moving under gravity alone, with air resistance negligible. It is the special case with zero acceleration along one axis and a constant, nonzero acceleration along the perpendicular one. With $+x$ horizontal and $+y$ up, and a launch at speed $v_0$ at angle $\theta$ above the horizontal:

The path is a parabola. Solve the $x$ equation for $t = \frac{x}{v_0\cos\theta}$ (launch from the origin) and substitute into $y$:

That is a downward-opening parabola in $x$. The shape depends on $a_y$ being constant; if the vertical acceleration varies, the path is a different curve.

At the top, the velocity is horizontal, not zero. The peak is where $v_y = 0$, at $t = \frac{v_0\sin\theta}{g}$. The horizontal component $v_0\cos\theta$ is unchanged, so the speed at the top is $v_0\cos\theta$, the minimum speed of the flight. The acceleration there is still $g$ downward.

Range on level ground. If the projectile lands at its launch height, set $y = 0$: $t\left(v_0\sin\theta - \frac{1}{2}gt\right) = 0$, so the flight time is $t = \frac{2v_0\sin\theta}{g}$. The horizontal distance covered is

The range is largest when $\sin 2\theta = 1$, at $\theta = 45^\circ$, and two angles adding to $90^\circ$ (such as $30^\circ$ and $60^\circ$) give equal ranges. This formula is only for equal launch and landing heights. On a hill, off a cliff, or at a wall, go back to the component equations.

The velocity graphs make the same story visible: $v_x$ against $t$ is a horizontal line, and $v_y$ against $t$ is a straight line of slope $-g$ that crosses zero at the peak. AP Physics C: Mechanics asks for quantitative work in two dimensions only; a third dimension adds a $z$ equation that behaves the same way.

Worked examples

Where students lose points

On the exam

Projectile questions are solved component by component, and graders look for the split: the horizontal equation with constant velocity, the vertical one with $a_y = -g$ (use $g = 10\ \text{m/s}^2$ unless told otherwise), and the time carried from one to the other. Expect launches from a height, landings on a raised surface, and requests to sketch $v_x$ and $v_y$ against time or to decide which of two launches stays up longer (the one with the larger $v_{y0}$, when both start and land at the same heights). When a position or velocity is given as a function of time, the question is testing calculus on each component, not projectile formulas.