Answer
$v_{max}=\sqrt{2gL(1-cos \theta_0)}$.
Work Step by Step
See Figure 11-12. When a pendulum of length L is released from rest, at an angle of $\theta_0$ away from the vertical, the mass m is at a height of $L(1-cos \theta_0)$ above the lowest point.
Use energy conservation to relate the potential energy at the starting, maximum height of the pendulum, and the kinetic energy at the lowest point. Let the lowest point represent zero gravitational potential energy.
$$PE_i=KE_f$$
$$mg L(1-cos \theta_0)=\frac{1}{2}mv_{max}^2$$
$$v_{max}=\sqrt{2gL(1-cos \theta_0)}$$
