How do i determine the landing point of a person falling from a building with a forward velocity of 2.2 fps at 45 degree angle at takeoff and a height of 50'

The components of velocity 
are Vh ( horizontal) = 2.2 cos 45  and vertical  Vv = 2.2 sin 45

to find the time when he lands we use

s = ut + 0.5at^2  where u = intial velocity  , s = height, t = time and a = acce;eration due to gravity.

so
50 = 2.2sin45 t + 0.5*32*t^2
50 = 1.5556 t + 16t^2

16t^2 + 1.5556t - 50 = 0
t = 1.72 seconds

horizontal distance travelled = velocity * t = vh * t

=  
2.2 cos 45 * 1.72  =  2.68 feet