First split up the velocity into horizontal and vertical components.
[tex]v_x = 13.9 cos (25) \\ \\ v_y = 13.9 sin (25)[/tex]
Next set up position functions for both 'x' and 'y' directions.
Using the general position function
[tex]\frac{a}{2} t^2 +vt + c[/tex]
a = acceleration, v = velocity, c = initial position
Assume initial position is at (0,0)
We know the velocities, vx and vy
acceleration due to gravity is -9.8 (Only in 'y' direction)
There is 0 acceleration in the 'x' direction.
[tex]x(t) = 13.9 cos (25) t \\ \\ y(t) =-\frac{9.8}{2} t^2 + 13.9 sin(25) t[/tex]
The football will hit the ground when y(t) = 0
[tex]-\frac{9.8}{2} t^2 + 13.9 sin(25) t = 0 \\ \\ -\frac{9.8}{2} t + 13.9 sin(25) = 0 \\ \\ t = \frac{(2)(13.9 sin (25))}{9.8}[/tex]
Substitute this value into x(t) to find how far the football traveled.
[tex]x(t) = 13.9 cos(25) (\frac{(2)13.9 sin(25)}{9.8}) = \frac{13.9^2 cos (25) sin(25)}{4.9}[/tex]
