Question

The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is

Answer 1

At maximum height
velocity = uCosθ
u/2 = uCosθ
θ = 60°

R = u² Sin(2θ) / g
= u² Sin(2 × 60°) / g
= u²√3 / (2g)

Range on horizontal plane is u²√3 / (2g)