jmc

Let $\ell$ the directrix of the parabola. Let $C$ and $D$ be the projections of $F$ and $B$ onto the directrix, respectively. Any point on the parabola is equidistant to the focus and the parabola, so $VF=VC$ and $BF=BD.$

Let $x=VF=VC.$ Then $BD=2x,$ so $BF=2x.$ By Pythagoras on right triangle $BFV,$ $$BV=\sqrt{V{F}^2+B{F}^2}=\sqrt{x^2+4x^2}=x\sqrt{5}.$$Then by the Law of Cosines on triangle $ABV,$ $$\cos \angle AVB=\frac{A{V}^2+B{V}^2-A{B}^2}{2\cdot AV\cdot BV}=\frac{5x^2+5x^2-16x^2}{2\cdot x\sqrt{5}\cdot x\sqrt{5}}=\boxed{-\frac{3}{5}}.$$