Answer
The potential energy is $U=m g L(1-\cos \theta)$ at the position shown in Fig. $8-34$ (which we consider to be the initial position). Thus, we have
$$
K_{i}+U_{i}=K_{f}+U_{f}
$$
$$
0+m g L(1-\cos \theta)=\frac{1}{2} m v^{2}+0
$$
$$
v=\sqrt{\frac{2 m g L(1-\cos \theta)}{m}}=\sqrt{2 g L(1-\cos \theta)}
$$
Plugging in $L=2.00 \mathrm{m}$ and $\theta=30.0^{\circ}$ we find $v=2.29 \mathrm{m} / \mathrm{s}$
Work Step by Step
The potential energy is $U=m g L(1-\cos \theta)$ at the position shown in Fig. $8-34$ (which we consider to be the initial position). Thus, we have
$$
K_{i}+U_{i}=K_{f}+U_{f}
$$
$$
0+m g L(1-\cos \theta)=\frac{1}{2} m v^{2}+0
$$
$$
v=\sqrt{\frac{2 m g L(1-\cos \theta)}{m}}=\sqrt{2 g L(1-\cos \theta)}
$$
Plugging in $L=2.00 \mathrm{m}$ and $\theta=30.0^{\circ}$ we find $v=2.29 \mathrm{m} / \mathrm{s}$
