Simple, isn't it?

@Aditya Kumar – I see now what I was doind wrong. The equation of the object is $-gx^2/(2v^2\cos(\alpha)^2)+x\sin(\alpha)/\cos(\alpha)+R=y$ where x=0 and y=0 are the coordinates of the center of the sphere. By replacing $v^2=gR/2$ we have $-x^2/(R\cos(\alpha)^2)+x\sin(\alpha)/\cos(\alpha)+R=y$ . If we replace $x=R\cos(\theta),y=R\sin(\theta)$ that represent the equation of the sphere then we have that $-R\cos(\theta)^2/\cos(\alpha)^2+R\cos(\theta)\sin(\alpha)/\cos(\alpha)+R=R\sin(\theta)$ The equation has two solutions: $\theta=90,\theta=\alpha$ . The first solution is when x=0 which we already know since is the starting point of the object. The second solution is the other point of intersection. The derivative of the previous equation for x has to be equal to the derivative of the equation of the sphere that is $-2x/(R\cos(\alpha)^2)+\sin(\alpha)/\cos(\alpha)=-x/\sqrt{R^2-x^2}$ . By replacing $x=Rcos(\alpha)$ we have: $-2/\cos(\alpha)+\sin(\alpha)/\cos(\alpha)=-\cos(\alpha)/\sin(\alpha)$

$-2+\sin(\alpha)=-\cos(\alpha)^2/\sin(\alpha)$ $-2=-\cos(\alpha)^2/\sin(\alpha)-\sin(\alpha)^2/\sin(\alpha)$

$2=1/\sin(\alpha)$ $\arcsin(1/2)=\alpha=30°$ . The only thing I am not sure of is why $v=\sqrt{gR/2}$ is the minimum velocity?