= Solution
{c}
Let $V>0$ be the observer's speed, so its laboratory velocity is $-V\mathbf e_x$. The particle's laboratory speed along $+y$ follows from the <energy–momentum relation>:
$$
u=c\sqrt{1-\frac{m^2c^4}{E^2}},\qquad q=\frac uc\in(0,1).
$$
Here $E>mc^2$ because the massive particle is moving. Boosting to the observer's frame with velocity $-V$ gives, by the <relativistic velocity-addition formula>,
$$
u'_x=V,\qquad u'_y=\frac u\gamma,\qquad \gamma=(1-V^2/c^2)^{-1/2}.
$$
Equivalently its <four-momentum> has $p'_x=\gamma VE/c^2$, $p'_y=p_y$. Thus $\tan\phi=u/(\gamma V)$. Solving this equation, rather than using the Galilean transverse velocity, yields
$$
\boxed{V=\frac{u}{\sqrt{\tan^2\phi+u^2/c^2}}=\frac{c q\cos\phi}{\sqrt{\sin^2\phi+q^2\cos^2\phi}}.}
$$
For a strictly moving observer, $0<\phi<\pi/2$. At $\phi=\pi/2$ the limiting observer speed is zero; as $\phi\downarrow0$ it tends to $c$. The direction of the observer's velocity remains negative $x$; the displayed $V$ is its positive magnitude.
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