Answer
The ball's speed at B is $6.55m/s$
Work Step by Step
Initially, at point A, the spring is compressed by $x_0=0.065m$, and the ball's speed $v_0=0$ and height $h_0=0$.
At point B, the ball's height $h_f=0.3m$ and has speed $v_f$. Since the spring is released, $x_f=0$
Since total energy is conserved, $$\frac{1}{2}mv_f^2+mgh_f=\frac{1}{2}kx_0^2$$ $$\frac{1}{2}mv_f^2=\frac{1}{2}kx_0^2-mgh_f$$ $$v_f=\sqrt{\frac{kx_0^2-2mgh_f}{m}}$$
We have $k=675N/m$, the ball's mass $m=0.0585kg$ and weight $mg=0.57N$. Therefore, the ball's speed at B is $$v_f=6.55m/s$$
