In projectile horizontal distance is given by :
[tex]h = \dfrac{u^2sin^ 2\theta}{2g}[/tex]
Putting all given values, we get :
[tex]h = \dfrac{u^2sin^ 2\theta}{2g}\\\\h=\dfrac{16^2\times sin^2 \ 55.3^o}{2\times 9.8}\\\\h=13.1\ m[/tex]
Therefore, maximum height of football is 13.1 m.
Hence, this is the required solution.