Question

If the velocity at the maximum height of a projectile is half its initial velocity of projection u, then find its range on the horizontal plane.

Answer 1

Given :

let u be the initial velocity of projection.

velocity at maximum height = Half of its initial velocity of projection 

uCosθ=u/2

cosθ=1/2

θ= 60° 

Range formula :

R=u²Sin2θ / g

=u²sin120°/ g

=√ 3/2 u²/ g

∴Range on the horizontal plane is √ 3/2 u²/ g