= Solution
In the parent's rest frame, the identical daughters have equal and opposite <momenta>, and each has total <energy> $110\,\mathrm{GeV}$. With daughter rest energy $Q=90\,\mathrm{GeV}$, each speed in this frame is
$$
\beta_*c=c\sqrt{1-\frac{Q^2}{110^2}},\qquad \beta_*=0.5749596.
$$
Transform to the rest frame of one daughter using the collinear <relativistic velocity addition> law. The other daughter then has speed
$$
\boxed{\frac wc=\frac{2\beta_*}{1+\beta_*^2}=\frac{44\sqrt{10}}{161}=0.864225.}
$$
An independent <invariant mass> check is useful. In the rest frame of one daughter the other's total <energy> satisfies $W^2=2Q^2+2QE'$, so $E'=178.8889\,\mathrm{GeV}$ and the relative <Lorentz factor> is $\gamma_{\rm rel}=E'/Q=161/81$. Hence $w/c=\sqrt{1-\gamma_{\rm rel}^{-2}}$, the same result.
The answer is unchanged if the parent is produced by a fixed-target collision. Its laboratory <velocity> then differs, but its rest <mass> and two-body decay kinematics do not. The scalar product of the daughters' <four-momenta>, equivalently their relative <Lorentz factor>, depends only on $W$ and $Q$. A common boost cannot change the speed of one measured in the rest frame of the other.<Relative speed from four-velocity contraction>[Invariant relative Lorentz factor and associated relative speed]{ref}
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