Question

A football.....

Answer 1

The maximum range occurs for a launch angle of 45°. At this angle, also the minimum velocity is needed. And you know that:
\[x = \frac{ V^2 \sin 2\alpha }{ g }\]then
\[V^2 = \frac{ xg }{ \sin 2\alpha }   \rightarrow   V^2 = xg   \rightarrow    V = \sqrt{xg} \]

Answer 2

THANKS!

Answer 3

;-)