A small block slides with velocity v0 = 0.5gr on the horizontal frictionless surface as shown in the fig. The blockleaves the surface at point C. The angle θ in the figure is:

At point C: mg cos θ -N =mv2r(when block leaves the surface normal force becomes zero, so putting N = 0) ⇒g cos θ = v2rv2 = v20+2gh, h = r - r cos θgr cos θ = (0.5gr)2+ 2g(r-r cos θ)⇒cos θ = 14+2-2 cos θ ⇒ cos θ = 34