3.1 Motion of a projectile

Ignoring air resistance, horizontal acceleration is zero and vertical acceleration is −g. With initial speed u at angle θ, x=u cosθ·t and y=u sinθ·t−½gt².

Treat the two components independently, then eliminate t or use symmetry. The launch and landing heights must be stated before using range or time-of-flight formulas.

For level ground, time of flight is 2u sinθ/g and range is u²sin2θ/g. The maximum range occurs at 45° only under this level-ground, no-drag model.

The velocity is not constant as a vector; only its horizontal component is constant, and gravity acts throughout the flight.

Eliminating time from the component equations gives y=x tanθ−gx²/(2u²cos²θ), a quadratic trajectory when gravity is uniform and air resistance is neglected.

Use the equation to find height, range or intersection with a target, but check that the chosen root corresponds to a future time and that the launch/landing geometry matches the question.

At a fixed horizontal distance, the quadratic may give two launch angles: a low path and a high path. Both can reach the point, but they have different flight times and maximum heights.

A parabolic path is an idealisation; drag, varying gravity or wind changes it, and an algebraic x-root is not automatically a physically valid time.

For a projectile launched with speed u at angle θ, x=u cosθ·t and y=u sinθ·t−½gt² still describe the motion. A target at a different height changes the time and range equations, not the component model.

Write the target condition in x and y, eliminate t, and solve only for values consistent with t≥0. Do not use the level-ground range formula unless launch and landing heights are equal.

A ball launched from a platform can hit a lower target on the descending path; the second root of the height equation represents a later intersection, while a negative time is discarded.

The 45° maximum-range result is not universal: it assumes equal heights, uniform gravity and no air resistance.