Kinematics — Quadratic Equations in Free Fall
Kinematics
When you throw a ball, drop your phone, or watch a diver plunge into a pool, the motion follows equations that are quadratic functions — parabolas on a graph.
For constant acceleration $a$, starting with velocity $v_0$ and position $x_0$:
$$x(t) = x_0 + v_0 t + \frac{1}{2}at^2$$
$$v(t) = v_0 + at$$
$$v^2 = v_0^2 + 2a(x - x_0)$$
The position equation is a quadratic — $x = \frac{1}{2}at^2 + v_0 t + x_0$ — the same form as $y = ax^2 + bx + c$ from algebra. The trajectory of every ball, bullet, and spacecraft follows a parabola.
Free Fall
Near Earth's surface, gravity provides $a = g = 9.8$ m/s² (downward). Drop an object from height $h$ with zero initial velocity:
$$h = \frac{1}{2}gt^2 \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g}}$$
You drop a ball from the top of a 45-meter building. How long does it take to hit the ground, and how fast is it going?
Time:
$$t = \sqrt{\frac{2 \times 45}{9.8}} = \sqrt{9.18} \approx 3.03 \text{ s}$$
Final velocity:
$$v = gt = 9.8 \times 3.03 \approx 29.7 \text{ m/s} \approx 107 \text{ km/h}$$
A ball dropped from 45 m hits the ground at highway speed in just 3 seconds. Square roots and multiplication give you the answer.
Projectile Motion — Trig Meets Quadratics
When you throw something at an angle $\theta$, the initial velocity splits into components using trigonometry:
$$v_x = v_0 \cos\theta, \qquad v_y = v_0 \sin\theta$$
The range of a projectile on flat ground is:
$$R = \frac{v_0^2 \sin(2\theta)}{g}$$
Maximum range occurs at $\theta = 45°$ because $\sin(90°) = 1$. This is why athletes throw at roughly 45°.
The kinematic equations are quadratic functions — parabolas. Every time you graph $y = ax^2 + bx + c$, you're plotting a possible trajectory of a thrown object. The square root, trig functions, and quadratic formula aren't abstract — they're how we predict every throw, jump, and fall.