← Back to lesson

The maximum-range angle (45°)

Free notes · JEE Main Physics

Formulas

sin⁡2θ=1⇒θ=45∘\sin 2\theta = 1 \Rightarrow \theta = 45^\circ

Condition for maximum range

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°