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Horizontal projection from a height

Free notes · JEE Main Physics

Formulas

T=2hg,R=u2hgT = \sqrt{\frac{2h}{g}},\quad R = u\sqrt{\frac{2h}{g}}

Horizontal projection

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°