The speed of a projectile when at its greatest height is 25  of its speed when at half of  its greatest height. Angle of projection is

Vx=ucosθ Vy=u2sin2θ−2gh Here Vy=u2sin2θ−2g(h2) =u2sin2θ−2g2(u2sin2θ2g) =u2sin2θ2 The velocity at that point beVm=Vx2+Vy2=u2cos2θ+u2sin2θ2 Velocity at heighest point is VH=Vx=ucosθ Given VH=25Vm VH2=25Vm2 u2cos2θ=25(2u2cos2θ+u2sin2θ2) tan2θ=3 ∴θ=600