Physics road embankment problem?

11m is the horizontal distance. The y component of the velocity is initially zero so y=1/2gt^2 solve for t, t=sqrt(2y/g)=sqrt(2*3/9.8)=0.612s.. In the x direction we have x=vt, v is what we what we want so v=x/t=11/0.612=18m/s