Answer with explanation:
Since the motion of the ball is projectile motion we shall use the equations for projectile motion.
The maximum height achieved by the projectile is given by
[tex]h_{max}=\frac{v_{o}^{2}sin^{2}(\theta )}{2g}[/tex]
Applying the values we get
[tex]h_{max}=\frac{12^{2}\times sin^{2}(45)}{2\times 9.81}\\\\h_{max}=3.669m[/tex]
The range of the projectile is given by
[tex]R=\frac{v_{o}^{2}sin(2\theta )}{2g}[/tex]
Applying values we get
[tex]R=\frac{12^{2}\times sin(2\times 45 )}{2\times 9.81}\\\\R=7.339m[/tex]
thus the maximum horizontal distance reached by the ball equals 7.339 meters after which it bounces thus a person standing 8 meters away will not be able to catch it.
(The height of the players is not taken into account since no info is given about their height.)
