Kinematics in Two Dimensions and Projectile Motion

Fire a bullet horizontally and drop another at the same instant: they hit the ground together. Gravity is completely indifferent to horizontal motion. Every projectile problem splits into two independent one-dimensional stories — constant velocity across, free fall down — sharing only the clock.

In 2D, position, velocity, and acceleration are vectors. The horizontal and vertical components are independent.

Position and velocity vectors

$$\vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j}, \qquad \vec{v}(t) = \frac{d\vec{r}}{dt}$$

Projectile motion — Launch at speed $v_0$ and angle $\theta$ above horizontal (neglecting air resistance):

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

Key results

Important principles

• The horizontal velocity $v_x = v_0\cos\theta$ is constant (no horizontal acceleration).
• The vertical motion is free fall with $a_y = -g$.
• Maximum range occurs at $\theta = 45^\circ$.
• Two angles $\theta$ and $90^\circ-\theta$ give the same range.

Physics insight: Galileo showed that the trajectory is a parabola by decomposing the motion into independent horizontal and vertical components.

Common pitfall: At the top of the arc the projectile is not momentarily "force-free". Its vertical velocity is zero for an instant, but gravity still acts at full strength — that is precisely why it comes back down.

Projectile Motion

A projectile launched with speed $v_0$ at angle $\theta$ follows a parabolic path. The horizontal and vertical motions are independent:

$$x = v_0 \cos\theta \cdot t, \quad y = v_0 \sin\theta \cdot t - \tfrac{1}{2}gt^2$$

• Range: $R = v_0^{2} \sin 2\theta / g$
• Max height: $H = v_0^{2} \sin^{2}\theta / (2g)$
• Time of flight: $T = 2v_0 \sin\theta / g$

Maximum range occurs at $\theta = 45^{\circ}$ (in vacuum). Air resistance reduces this angle in practice.