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Projectile Motion: Trajectory, Time of Flight, Range

Free notes · JEE Main Physics

Formulas

x=v0cos⁡θ t,y=v0sin⁡θ t−12gt2x = v_0 \cos\theta \, t, \qquad y = v_0 \sin\theta \, t - \tfrac{1}{2} g t^2

Position of a projectile at time t (origin at launch point)

y=xtan⁡θ−gx22v02cos⁡2θy = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}

Trajectory equation — y as a function of x (a parabola)

T=2v0sin⁡θg,H=v02sin⁡2θ2g,R=v02sin⁡2θgT = \frac{2 v_0 \sin\theta}{g}, \qquad H = \frac{v_0^2 \sin^2\theta}{2g}, \qquad R = \frac{v_0^2 \sin 2\theta}{g}

Time of flight T, maximum height H, and horizontal range R

Key points

Worked example

A projectile is launched at 40 m/s at 30° above horizontal. Take g = 10 m/s². Find (a) time of flight, (b) maximum height, (c) range.

  1. T = 2 v0 sinθ / g = 2 × 40 × sin 30° / 10 = 2 × 40 × 0.5 / 10 = 4 s
  2. H = v0² sin²θ / 2g = 1600 × 0.25 / 20 = 20 m
  3. R = v0² sin 2θ / g = 1600 × sin 60° / 10 = 1600 × 0.866 / 10 ≈ 138.6 m

Answer: T = 4 s, H = 20 m, R ≈ 138.6 m