2: Motion in 2D (vectors, projectile motion)

Motion in 2D: Vectors and Projectile Motion

Kinematics extends naturally into two dimensions, requiring vectors to describe motion fully. Unlike 1D motion confined to a straight line, 2D motion occurs in a plane, demanding direction alongside magnitude. Vectors provide this essential directional information. Key vector quantities include position ($\vec{r}$), displacement ($\Delta \vec{r} = \vec{r_f} - \vec{r_i}$), velocity ($\vec{v} = d\vec{r}/dt$), and acceleration ($\vec{a} = d\vec{v}/dt$). These vectors are often resolved into perpendicular components, typically using Cartesian coordinates (x and y axes). For example, $\vec{v} = v_x \hat{i} + v_y \hat{j}$. The magnitude and direction of any vector are found using Pythagoras ($v = \sqrt{v_x^2 + v_y^2}$) and trigonometry ($\theta = \tan^{-1}(v_y / v_x)$).

Projectile motion is a quintessential application of 2D kinematics, describing the trajectory of an object launched with an initial velocity under constant acceleration due to gravity (neglecting air resistance). The key insight is decomposing the motion into independent horizontal (x) and vertical (y) components:

1. Horizontal Motion (x-direction): Acceleration is zero ($a_x = 0$). The horizontal velocity remains constant ($v_x = v_{0x} = v_0 \cos \theta_0$, where $\theta_0$ is the launch angle). The horizontal position is given by $x = x_0 + v_{0x} t$.
2. Vertical Motion (y-direction): Acceleration is constant downward ($a_y = -g$, where $g \approx 9.8 \, \text{m/s}^2$). The vertical velocity changes with time ($v_y = v_{0y} - gt = v_0 \sin \theta_0 - gt$). The vertical position is $y = y_0 + v_{0y} t - \frac{1}{2} g t^2$.

The combination of constant horizontal velocity and uniformly accelerated vertical motion produces a parabolic trajectory. Critical characteristics include:

• Time of Flight: The total time the projectile is airborne. For launch and landing at the same height ($y = y_0$), it's $T = \frac{2 v_0 \sin \theta_0}{g}$.
• Maximum Height: The peak of the trajectory, reached when $v_y = 0$. $H = \frac{(v_0 \sin \theta_0)^2}{2g}$.
• Range: The horizontal distance traveled when landing at the launch height ($y = y_0$). $R = \frac{v_0^2 \sin(2\theta_0)}{g}$, showing maximum range at $\theta_0 = 45^\circ$.

Solving projectile problems consistently involves identifying knowns (initial position, initial velocity components or magnitude/angle, time, positions), applying the independent kinematic equations for x and y, and finding connections (often time) between the motions. Vector addition/subtraction and component resolution are fundamental skills throughout.