Answer
The ram's speed at the point of contact is $13.8m/s$
Work Step by Step
Before being released, the ram spring is compressed by $y_0=3\times10^{-2}m$. At this point, the ram's speed $v_0=0$ and height $h_0=0$.
After being released, the ram spring is compressed by $y_f=0.8\times10^{-2}m$. At this point, the ram's speed $v_f$ and height $h_f=y_f-y_0=-2.2\times10^{-2}m$.
Since total energy is conserved, $$\frac{1}{2}mv_f^2+mgh_f+\frac{1}{2}ky_f^2=\frac{1}{2}ky_0^2$$ $$\frac{1}{2}mv_f^2=\frac{1}{2}ky_0^2-mgh_f-\frac{1}{2}ky_f^2$$ $$v_f=\sqrt{\frac{k(y_0^2-y_f^2)-2mgh_f}{m}}$$
We have $k=3.2\times10^4N/m$, the ram's mass $m=0.14kg$ and weight $mg=1.37N$. Therefore, the ram's speed at the point of contact is $$v_f=13.8m/s$$
