= Solution
Use coordinates $s$ down the <inclined plane> and $n$ along its upward normal, with the launch point as origin. Gravity has components $g\sin\phi$ and $-g\cos\phi$. For launch angle $\theta$ above the downhill direction,
$$
s(t)=v_0\cos\theta\,t+\frac12g\sin\phi\,t^2,
\qquad n(t)=v_0\sin\theta\,t-\frac12g\cos\phi\,t^2.
$$
The nonzero root of $n(t)=0$ is $t_f=2v_0\sin\theta/(g\cos\phi)$. Substituting into $s(t)$ and combining the two trigonometric terms gives the distance along the slope:
$$
\boxed{R_{\rm slope}=\frac{2v_0^2\sin\theta\cos(\theta-\phi)}{g\cos^2\phi}.}
$$
This assumes $0\le\phi<\pi/2$, $\sin\theta>0$, and a downhill landing with $\cos(\theta-\phi)>0$, as well as the neglected <air resistance>. It reduces to the horizontal <projectile range> $v_0^2\sin(2\theta)/g$ at $\phi=0$. The angle relative to the horizontal is $\theta-\phi$, which explains the shifted cosine. The original diagram uses stipulated illustrative angles; no examination figure is being reconstructed.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014-resit-projectile-on-slope.png]
{title=Original <projectile motion> geometry with the launch angle measured above a downhill <inclined plane>}
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