EASA/GCAA Part-66 · Module 2 — Physics · Mechanics

Projectile Motion

Objective: a projectile's horizontal and vertical motions are independent. Horizontal velocity stays constant; vertical velocity changes under gravity. Together they make a parabola. Find the 45° launch angle that gives maximum range on level ground.

Trajectory — with velocity components

Launch speed 30 m/s

Launch angle 45°

Gravity g 9.81 m/s²

Range

Max height

Time of flight

Vx · Vy now

0 · 0

Range (level, no drag): R = v²·sin(2θ) / g · Max height: H = (v·sinθ)² / 2g

Split the launch velocity into a horizontal part (v·cosθ) and a vertical part (v·sinθ). The horizontal velocity never changes — there is no horizontal force in a vacuum. The vertical velocity decreases going up, reaches zero at the apex, then increases coming down, exactly as if the object were thrown straight up. Because range depends on sin(2θ), it is greatest at θ = 45°, and complementary angles (e.g. 30° and 60°) give the same range. Real air resistance shortens and un-symmetrises the arc.

Try this: Set 45° and fire, then try 30° and 60° — same range, different shape.
Try this: Watch the green vertical vector shrink to zero at the top, while the blue horizontal vector never changes.
Try this: Switch on drag — the range falls and the descent steepens.
Try this: Lower gravity and see how much farther and higher the projectile travels.