Answer
$$
\begin{aligned} v =8.35 \mathrm{\ m} / \mathrm{s} \end{aligned}
$$
Work Step by Step
Careful examination of the figure leads to the trigonometric relation $$h=L-L \cos \theta$$ when the angle is measured from vertical as shown. Thus, the gravitational potential energy is $$U=m g L\left(1-\cos \theta_{0}\right)$$ at the position shown in Fig. $8-34$ (the initial position). Thus, we have
$$
K_{0}+U_{0}=K_{f}+U_{f}
$$
$$
\frac{1}{2} m v_{0}^{2}+m g L \bigcup_{1-\cos \theta_{0}} d=\frac{1}{2} m v^{2}+0
$$
which leads to
$$
\begin{aligned} v &=\sqrt{\frac{2}{m}\left[\frac{1}{2} m v_{0}^{2}+m g L\left(1-\cos \theta_{0}\right)\right.}]=\sqrt{v_{0}^{2}+2 g L\left(1-\cos \theta_{0}\right)} \\ &=\sqrt{(8.00 \mathrm{m} / \mathrm{s})^{2}+2\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)(1.25 \mathrm{m})\left(1-\cos 40^{\circ}\right)}=8.35 \mathrm{\ m} / \mathrm{s} \end{aligned}
$$
