Projectiles

Two motions, one clock

Launching at speed u and angle θ gives u cos θ across and u sin θ up. With air resistance modelled away, nothing horizontal ever changes the horizontal speed, and nothing vertical is anything but gravity. The two motions run independently and are joined only by the shared time t.

The standard questions

Greatest height is a vertical question, with v = 0 upwards giving h = (u sin θ)²/(2g). Landing on level ground means the vertical displacement is zero, which is a different statement from the velocity being zero. A projectile launched horizontally is the same model with u sin θ = 0, starting at its highest point.

For the velocity partway through, rebuild the components. The horizontal one has not changed. The vertical one is u sin θ − gt, negative once the ball is falling. For the golf ball at t = 1.5 s, the components are 17.3 across and −4.7 vertically, giving speed √(17.3² + 4.7²) ≈ 17.9 m s⁻¹ at tan⁻¹(4.7/17.3) ≈ 15.2° below the horizontal.

The equation of the path

Eliminating t between the two directions gives the shape of the whole flight instead of one instant of it. Horizontally x = ut cos θ, so t = x/(u cos θ). Substituting into y = ut sin θ − ½gt² gives

which is a quadratic in x, so the path is a parabola. It sits in neither camp. The booklet does not print it, and it is not one of the results you are told to know, because the specification expects you to derive it. So derive it, every time, from the two component equations. Examiners often want it in the form with sec²θ or with (1 + tan²θ), and the derivation itself is the marked part. Show the substitution.

As a worked instance, a ball projected at 20 m s⁻¹ and 30° is at y = 20 tan 30° − 9.8 × 20²/(2 × 20² × cos²30°) = 11.55 − 6.53 = 5.01 m when it has travelled 20 m horizontally. That matches the t = 1.15 s answer from the two-equation route, as it must.