So the initial velocity has an x- and y-component because it's a vector. The x-component of velocity remains the same throughout the kick. If there was no ground for it to land on, it would continue on forever.
The y-component of velocity obviously changes because we're dealing with acceleration due to gravity.
So I'm going to look at the time for the ball to get to the top. At the top, final y-velocity is 0.
So: Vi = 8sin(40), Vf = 0, a = -9.81, and we're looking for t
V = Vi + at
0 = 8sin(40) -9.81(t)
t = 0.52419 s
This is the time it takes to get to the top. I could easily take a shortcut and double that time (because of symmetry), but I won't because you might have problems in the future that have different initial- and final y-positions. If you tried doubling it in that case, you'd end up with the wrong answers, so I'll take the long way for now.
Okay so now we need to find the height at t=0.52419 s:
y = y_o + v_o*t +.5at^2
y = 0 + (8sin(40))*(0.52419) + (0.5)(-9.81)(0.52419)^2
y = 1.35 m
Now we can find the final y-velocity:
v^2 = v_o^2 -2a(y-y_o)
v^2 = 0 - 2(-9.81)(1.35-0)
v = sqrt (26.487) = -5.147 m/s (minus because it's going down)
Now we can find the time from y = 1.35 m to y=0 m (where it lands)
y-y_o = ((v +v_o)/2)*t
1.35 - 0 = ((-5.147 +0)/2)*t
t = 0.5251 s (I must have some rounding error here)
So the total time it takes to get from the initial position to the final is 0.52419 s + 0.5251 s = 1.050s
Now, we know the total time, we know that the x-component of the initial velocity remains constant. With this, we can calculate the total distance using the first equation:
http://www.rdoman.com/phy/motf/motdrw01.gif
(8cos(40))*(1.050s) = 6.435 m