Answer:
For Ryan:
[tex]at \: highest \: point : v = 0 \\ velocity = 0 \: {ms}^{ - 1} [/tex]
For the football kicked:
[tex]s = ut - \frac{1}{2} g {t}^{2} \\ for \: range : s = 0 \\ u \sin( \theta) = \frac{gt}{2} \\ t = \frac{2u \sin(\theta) }{g} \\ range(x) = ut \\ x = \frac{ {2u {}^{2} \sin(\theta) }^{} }{g} \\ \frac{50 \times 9.8}{ {2318.9}^{} } = \sin(\theta) \\ \\but \: 0 = u \sin(25) - 2 \times 9.8 \times 50 \\ u = 2318.9 \: {ms}^{ - 1} \\ \theta = 12.2 \degree[/tex]
