In horizontal level ground to ground projectile if at any instant velocity becomes perpendicular to initial velocity then what can you say about projection angle with horizontal.

Velocity at any time v→ = u→+g→t⇒ v→ = u cosθi⏞+(u sinθ-gt)j⏞Let at any time this velocity becomes perpendicular to initial velocity. The v→.u→ = 0Solve to get t = ugsinθNow t should be less than/eqaul to time of flight. So t ≤ Tug sinθ≤2u sinθg        ⇒ sin2θ ≥ 12⇒ sin θ ≥ 12    ⇒ θ ≥ 450