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Maximum height

Free notes · JEE Main Physics

Formulas

H=u2sin2θ2gH = \frac{u^2\sin^2\theta}{2g}

Maximum height

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°