Question

the velocity at the maximum height of a projectile is half of its initial velocity of projection. the angle of projection is

Answer 1

As ucosθ0=u/2ucos⁡θ0=u/2∴θo=60∘∴θo=60∘So R=u2sin(2×60∘)gR=u2sin⁡(2×60∘)g=3–√2u2g

Answer 2

At maximum height velocty = uCosθ

uCosθ = u/2
Cosθ = 1/2
θ = 60°

Angle of projection is 60°