Solution
= Solution
The rod imposes a <mechanical constraint>, and its force is perpendicular to the wrecking ball's instantaneous <velocity>, so it does no <work>. In the constant parallel <Newtonian gravitational field>, the ball falls through the vertical distance $l-l\cos\theta_0$. <Conservation of energy> between $B$ and $W$ therefore gives
$$
\frac12mv_1^2=mgl(1-\cos\theta_0),
$$
and hence
$$
\boxed{v_1=\sqrt{2gl(1-\cos\theta_0)}}.
$$