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The trajectory equation

Free notes · JEE Main Physics

Formulas

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}

Trajectory of a projectile

Worked example

For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

  1. R = v0² sin 2θ / g
  2. R is maximum when sin 2θ = 1, i.e. 2θ = 90°, so θ = 45°

Answer: 45°