= Solution
Let $u>0$ be the projectile's final <speed>, taking its deflection above the $x$-axis. <Momentum conservation> gives
$$
\mathbf v_1=u(\cos30^\circ,\sin30^\circ),\qquad
\mathbf v_2=\frac12\bigl(V-u\cos30^\circ,-u\sin30^\circ\bigr).
$$
Substituting these into <kinetic energy conservation>, $V^2=u^2+2|\mathbf v_2|^2$, gives $3u^2-2Vu\cos30^\circ-V^2=0$. Its positive root is $u=V(\sqrt3+\sqrt{15})/6$. Hence
$$
\boxed{\mathbf v_1=V\left(\frac{1+\sqrt5}{4},\frac{\sqrt3+\sqrt{15}}{12}\right),\qquad
\mathbf v_2=V\left(\frac{3-\sqrt5}{8},-\frac{\sqrt3+\sqrt{15}}{24}\right).}
$$
These vectors satisfy $\mathbf v_1+2\mathbf v_2=(V,0)$ and $|\mathbf v_1|^2+2|\mathbf v_2|^2=V^2$. For the reflected deflection, reverse both $y$ components.
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