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Every time a basketball arcs toward the hoop or a firework bursts into the night sky, its path traces a mathematically perfect curve called a parabola.
This happens because gravity pulls downward with constant acceleration while the horizontal speed remains steady, splitting two-dimensional motion into two independent parts.
In 1638, Galileo Galilei proved this independence across orthogonal dimensions: projectile motion decouples into uniform horizontal motion (ax=0a_x = 0) and uniformly accelerated vertical motion governed solely by gravity (ay=βˆ’g=βˆ’9.8Β m/s2a_y = -g = -9.8\text{ m/s}^2).
By analyzing these orthogonal dimensions independently, we decouple the 2D path into uniform horizontal motion with zero acceleration (ax=0a_x = 0) and vertical motion governed solely by gravity (ay=βˆ’g=βˆ’9.8Β m/s2a_y = -g = -9.8\text{ m/s}^2).
How do we write a single mathematical rule that predicts the projectile's height at every fraction of a second?

The Vertical Position Equation

When an object moves with constant acceleration aa, its displacement changes quadratically with time according to the kinematic equation Ξ”y=v0yt+12ayt2\Delta y = v_{0y}t + \frac{1}{2}a_y t^2.
Because gravity acts downward with acceleration gβ‰ˆ9.8Β m/s2g \approx 9.8\text{ m/s}^2, we set ay=βˆ’ga_y = -g, giving the standard vertical position equation:
y(t)=βˆ’12gt2+v0yt+y0y(t) = -\frac{1}{2}gt^2 + v_{0y}t + y_0
Here, y0y_0 is the initial height, v0yv_{0y} is the initial vertical velocity, and tt is time in seconds.
Because the leading coefficient a=βˆ’12ga = -\frac{1}{2}g is negative, the graph of y(t)y(t) is always a parabola that opens downward, guaranteeing a single peak maximum.
How do we find the exact instant the projectile reaches this highest point, and how high does it go?