Answer
The ball leaves the club at $v=55m/s$
Work Step by Step
1) Find final kinetic energy $KE_f$
According to work-energy theorem, $KE_f=KE_0+W (1)$
We have $m_{ball}=0.045kg$ and $v_0=0$, so $$KE_0=\frac{1}{2}mv^2_0=0$$
We also have $\sum F=6800N$ and $s=0.01m$; since the net force acts parallel to the ball's motion, $\theta=0$. The work done by the force on the ball is $$W=(\sum F\cos0)s=68J$$
Therefore, from (1), $$KE_f=68J$$
2) We know that $$KE_f=\frac{1}{2}mv_f^2=68$$ $$v^2_f=\frac{2\times68}{m}=3022.222m^2/s^2$$ $$v_f=55m/s$$
