Projectile Motion

Split any projectile's motion into x and y components the moment it leaves the launcher. Horizontally, nothing pushes or pulls it (ignoring air resistance), so v_x stays constant the whole flight: x(t) = v0 cos(theta)·t. Vertically, gravity accelerates it downward at g, so v_y changes steadily: y(t) = v0 sin(theta)·t - (1/2) g t^2. These two motions don't interact; gravity never touches v_x, and nothing touches v_y except gravity.

The parabola shape comes directly from this split. Solve the x equation for t = x / (v0 cos theta), then substitute into the y equation. The t² term becomes an x² term, giving y as a function of x that looks like y = ax - bx² — exactly the form of a parabola. That's why a thrown football, a cannonball, or a kicked soccer ball always traces the same family of curve: one coordinate grows linearly in time, the other quadratically, and eliminating time mixes them into x^2.

This is where derivatives and integrals from earlier in the unit pay off directly: v_x and v_y are just the time-derivatives of x(t) and y(t), and acceleration is the derivative of velocity — here, a_x = 0 and a_y = -g at every instant, integrated back to get the velocity and position functions. The range formula, R = v0² sin(2 theta)/g, falls out of setting y = 0 and solving for x. Landing below launch height breaks this symmetric formula, since the vertical drop takes longer than the symmetric parabola assumes, giving a longer flight time and a longer range than the flat-ground case at the same angle.