Answer
$$v_{0}=4.33 \mathrm{\ m} / \mathrm{s}$$
Work Step by Step
We look for the initial speed required to barely reach the horizontal position described by $$v_{h}=0 \quad and \quad \theta=90^{\circ}$$ (or $\theta=-90^{\circ},$ if one prefers, but since $\cos (-\phi)=\cos \phi,$ the sign of the angle is not a concern).
$$
K_{0}+U_{0}=K_{h}+U_{h}
$$
$$
\frac{1}{2} m v_{0}^{2}+m g L \quad b_{1-} \cos \theta_{0} y=0+m g L
$$
$v_{0}=\sqrt{2 g L \cos \theta_{0}}=\sqrt{2\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)(1.25 \mathrm{m}) \cos 40^{\circ}}=4.33 \mathrm{\ m} / \mathrm{s}$
