The Timpson Lyceum · University Physics I (2025)

Motion in Two or Three Dimensions – Lecture 3 (University Physics I)

Overview

This lecture extends one-dimensional kinematics to two- and three-dimensional motion, using vectors and the trick of splitting motion into independent components. It covers two headline applications: projectile motion (an object moving under gravity alone, which always traces a downward parabola) and uniform circular motion (motion around a circle at constant speed). Along the way it builds the angular language — radians, angular velocity, period, frequency — and derives centripetal acceleration. The recurring theme: a hard 2-D problem becomes two easy 1-D problems once you resolve it into perpendicular components.

Key Concepts

Projectiles. A projectile is an object on which the only force is gravity (air resistance ignored). Its motion stays in a single vertical plane, so it is a 2-D problem with $a_x = 0$ and $a_y = -g$ ( $g = 9.8\ \text{m/s}^2$ ). Resolve the launch velocity by angle: $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$ . The horizontal and vertical motions are then completely independent — constant velocity across, free fall up/down — and (taking the launch point as the origin):

\begin{aligned} x &= (v_0\cos\theta)\,t \\ y &= (v_0\sin\theta)\,t - \tfrac{1}{2} g t^{2} \\ v_x &= v_0\cos\theta \\ v_y &= v_0\sin\theta - g t \end{aligned}

Trajectory and its two key results. Eliminating $t$ between the $x$ and $y$ equations gives $y$ as a quadratic in $x$ — the path is always a downward-opening parabola. At the peak the vertical velocity is zero ( $v_y = 0$ ); using that and the symmetry of the parabola yields the maximum height $H$ and the horizontal range $D$ :

\begin{aligned} H &= \frac{v_0^{2}\sin^{2}\theta}{2g} \\ D &= \frac{v_0^{2}\sin 2\theta}{g} \end{aligned}

These pass the intuition check: more launch speed → higher and farther; more gravity → lower and shorter; height is greatest at $\theta = 90^\circ$ (straight up), and range is greatest at $\theta = 45^\circ$ (since $\sin 2\theta$ peaks there).

The monkey-and-hunter problem. Aim a projectile directly at a target that drops the instant you fire. You always hit it — regardless of distance, launch angle, or launch speed — because the projectile and the target fall by the same $\tfrac{1}{2} g t^{2}$ in the same time, so gravity removes exactly the same amount from both. (The only caveat: the projectile must reach the target before hitting the ground.)

Radians and angular quantities. An angle in radians is arc length over radius, so it is dimensionless; the arc length for angle $\theta$ is $s = r\theta$ . Define angular displacement $\Delta\theta$ , angular speed $\omega = \Delta\theta/\Delta t$ (rad/s), and angular acceleration $\alpha = \Delta\omega/\Delta t$ — the exact rotational analogues of $x$ , $v$ , $a$ . They connect to the linear quantities through the radius:

\begin{aligned} s &= r\theta \\ v &= \omega r \\ a_{\text{rad}} &= \omega^{2} r = \frac{v^{2}}{r} \end{aligned}

Period and frequency. The period $T$ is the time for one full revolution and the frequency $f$ is revolutions per second; they are reciprocals, with $f$ measured in hertz ( $\text{s}^{-1}$ ):

$T = \frac{2\pi}{\omega}, \qquad f = \frac{1}{T}$

Uniform circular motion. "Uniform" means $\alpha = 0$ , so $\omega$ — and therefore the speed $|\vec v| = \omega r$ — is constant. The velocity is not constant, though: its direction is always changing. That changing direction is an acceleration directed radially inward (centripetal), $a_{\text{rad}} = v^{2}/r$ . Faster spin or a tighter circle means more centripetal acceleration.

Worked Example

A projectile launched at $v_0 = 20\ \text{m/s}$ , angle $\theta = 30^\circ$ above horizontal. Find its maximum height and horizontal range ( $g = 9.8\ \text{m/s}^2$ ).

Maximum height — apply the height result, then substitute:

\begin{aligned} H &= \frac{v_0^{2}\sin^{2}\theta}{2g} \\ &= \frac{(20)^2\,\sin^{2}30^\circ}{2(9.8)} \approx 5.1\ \text{m} \end{aligned}

Range — apply the range result, then substitute:

\begin{aligned} D &= \frac{v_0^{2}\sin 2\theta}{g} \\ &= \frac{(20)^2\,\sin 60^\circ}{9.8} \approx 35.3\ \text{m} \end{aligned}

As a check, both grow with $v_0$ and shrink with $g$ , as expected; and had the launch angle been $45^\circ$ , the range would have been at its maximum.

Summary

Resolve 2-D motion into independent perpendicular components — that turns one hard problem into two simple 1-D ones.
A projectile has $a_x = 0$ , $a_y = -g$ ; its path is always a downward parabola.
Maximum height $H = \dfrac{v_0^{2}\sin^{2}\theta}{2g}$ (peak at $90^\circ$ ); range $D = \dfrac{v_0^{2}\sin 2\theta}{g}$ (max at $45^\circ$ ).
Angular quantities mirror linear ones and link through the radius: $s = r\theta$ , $v = \omega r$ , with period $T = 2\pi/\omega$ and frequency $f = 1/T$ .
In uniform circular motion the speed is constant but the velocity is not; the inward (centripetal) acceleration is $a_{\text{rad}} = v^{2}/r$ .