Verified answer

The motion of football consists in a horizontal component of uniform motion:

x=vx t where vx=35 cos(37°)=27.95 m/sec

and a vertical component of accelerated motion

y=-1/2gt^2+vyt+y0

y0=0

g=9.81 m/s^2

vy=35 sin(37)=21.06 m/s

When the football arrived on the ground y=0

0=-0.5 9.81 t^2+21.06t

4.905t^2-21.06t=0 second degree equation without known term

t(4.905t-21.06)=0

the first solution is t=0s. It is impossible!

the second solution is t=21.06/4.905; t=4,29 sec time spent in air from the football

For the horizontal distance, you can take x=vx t:

x=27.95 4,29

x=119,90 m

Wouldn't you also need the weight of the ball or something?

Or do you just need to assume the weight?

Or am I stupid beyond belief?

Oh, well.