Question

The magnitude of the velocity of projection being fixed, how does the range of projectile depends upon its angel of projection? Is there any optimum value for the angle of projection so that range may be a maximum?

Answer 1

The horizontal range of projectile is given by

R = u² sin 2θ/g

As the magnitude of the velocity of projection is fixed and g is constant at a given place, therefore R ∝ sin 2θ.

For sin 2θ = 1, i.e., θ = 45°, the range of the projectile is maximum.