y = y0 +v0*t +0.5at^2
where y0 = initial vertical position = 22m
y = final vertical position = 0m
v0 = initial vertical velocity = 0 m/s
a = acceleration = -9.8 m/s^2
t = time in seconds
0 = 22 +0*t + 0.5(-9.8)t^2
t^2 = 22/4.9 = 4.49 s^2
t = 2.12 s
So it traveled 35m in 2.12 s
the horizontal distance traveled is determined by:
x = x0 +v0*t +0.5at^2
but here a in the horizontal direction is 0 m/s^2
and v0 is in the velocity in the horizontal direction in this equation
35 m = 0 +v0*t
35 m = v0(2.12 s)
v0 = 16.5 m/s
So the ball was kicked 16.5 m/s in the horizontal direction