Start with 2 equations, one for the vertical distance and one for the horizontal distance.
s(vertical) = ut + 1/2at^2, but u = 0 as its projected horizontally so s(vertical) = 1/2(9.8)(t^2)
s(horizontal) = ut + 1/2at^2, but there is no horizontal acceleration so simply s(horiz) = ut, where u = 16
Also, the horizontal component velocity never changes, i.e. it is always 16 as there is no acceleration. The vertical component will change because of acceleration due to gravity.