= Solution
Use the standard meaning of <static spacetime>: a <stationary spacetime> chart whose time-space components also vanish. The printed static definition needs to inherit the timelike requirement $g_{00}<0$; without it, the requested timelike conclusion does not follow.
The preceding argument shows that $V=\partial_0$ is a timelike <Killing vector field>. Its metric-dual <one-form> is $v=V_\mu dx^\mu=g_{00}dx^0$. Its <exterior derivative> satisfies
$$
v\wedge dv=g_{00}dx^0\wedge (dg_{00}\wedge dx^0)=0.
$$
For a <torsion-free connection>, antisymmetrizing the <covariant derivative> of a <one-form> gives its <exterior derivative>. Consequently \b[the Killing vector has hypersurface orthogonality]:
$$
\boxed{V_{[\alpha}\nabla_\mu V_{\nu]}=0.}
$$
Geometrically, the hypersurfaces $x^0=\mathrm{constant}$ are orthogonal to $V$.
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