Answer
$$
v_{0}=7.45 \mathrm{m} / \mathrm{s}
$$
Work Step by Step
For the cord to remain straight, then the centripetal force (at the top) must be (at least) equal to gravitational force:
$$
\frac{m v_{t}^{2}}{r}=m g \Rightarrow m v_{t}^{2}=m g L
$$
where we recognize that $r=L .$ We plug this into the expression for the kinetic energy (at $\left.\text { the top, where } \theta=180^{\circ}\right) .$
$$
K_{0}+U_{0}=K_{t}+U_{t}
$$
$$\frac{1}{2} m v_{0}^{2}+m g L\ [{1-\cos \theta_{0}} ]=\frac{1}{2} m v_{t}^{2}+m g [ 1-\cos 180^{\circ}]$$
$$\frac{1}{2} m v_{0}^{2}+m g L \quad [{1-\cos \theta_{0}} ]=\frac{1}{2}(m g L)+m g(2 L)$$
which leads to
$$
v_{0}=\sqrt{g L\left(3+2 \cos \theta_{0}\right)}=\sqrt{\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)(1.25 \mathrm{m})\left(3+2 \cos 40^{\circ}\right)}=7.45 \mathrm{m} / \mathrm{s}
$$
