Answer
$v_f=45.93m/s$
Work Step by Step
According to the work-energy theorem, $$E_f-E_0=W_{nc}$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=W_{nc}$$
We take the initial point to be when the bat hits the ball and the final point to be 25m above the point of impact.
So $v_0=40m/s$, $h_f-h_0=25m$, $m=0.14kg$ and $W_{nc}=70J$. Therefore, $$0.07(v_f^2-1600)+1.372(25)=70$$ $$0.07(v_f^2-1600)=35.7$$ $$v_f^2=2110$$ $$v_f=45.93m/s$$
