Answer:
The final velocity is 4.71 m/s
Explanation:
The potential energy corresponding to a body of mass (m) at a height (h) is given as:
[tex]PE = Mgh-----(1)[/tex]
where g = acceleration due to gravity = 9.8 m/s2
The kinetic energy of a body of mass (m) moving at a velocity (v) is:
[tex]KE = 1/2Mv^{2} ---------(2)[/tex]
It is given that:
the height (h) from which the basketball hits the ground = 1.13 m
Based on equation (1), the PE is:
[tex]PE = M*9.8ms^{-2} *1.13 m = M*11.1 m^{2}s^{-2}[/tex]
During its descent, all the potential energy gets converted to kinetic energy
[tex]PE = KE\\\\M*11.1 m^{2} s^{-2} = \frac{1}{2} *M*v^{2} \\\\v = 4.71 m/s[/tex]
