= Solution
{c}
In terms of the positive <cylindrical radius> $r=-z$, the conserved <mechanical energy> becomes
$$
(1+\sqrt2)\dot r^2+U(r)=E,\qquad
U(r)=gr+\frac{a^2v^2}{2r^2},\qquad E=ga+\frac{v^2}{2}.
$$
The <effective potential> $U$ tends to infinity at both ends of $(0,\infty)$ for $v>0$, and $U''(r)=3a^2v^2/r^4>0$. Thus its two <turning points> enclose the allowed radial motion. At a <turning point>, $\dot r=0$, so
$$
2gr^3-(v^2+2ga)r^2+a^2v^2
=(r-a)(2gr^2-v^2r-av^2)=0.
$$
Besides $r=a$, the positive <root of a polynomial> is
$$
r_*=\frac{v^2+\sqrt{v^4+8gav^2}}{4g};
$$
the other <root of a polynomial> is negative. The <quadratic polynomial> $Q(r)=2gr^2-v^2r-av^2$ is negative at zero and has just one positive <root of a polynomial>. Since $Q(a)=2a(ga-v^2)$, we have $r_*<a$ when $v^2<ga$, and $r_*>a$ when $v^2>ga$.
Height is $z=-r$, so its <maximum> occurs at the smaller positive <turning point>. Hence
$$
\boxed{z_{\max}=\begin{cases}
-\dfrac{v^2+\sqrt{v^4+8gav^2}}{4g},&0<v<\sqrt{ga},\\[6pt]
-a,&v>\sqrt{ga}.
\end{cases}}
$$
Indeed, the initial vertical <acceleration> is
$$
\ddot z(0)=\frac{g-v^2/a}{2(1+\sqrt2)}.
$$
Below the critical speed the particle first rises to $z=-r_*$; above it the particle initially descends, and its initial height is already its greatest height. At $v=\sqrt{ga}$ the <turning points> merge and the particle follows a horizontal <circular orbit> at $z=-a$. For zero initial speed the radial <conservation of energy> gives motion to the apex level $z=0$; if the hole excludes the apex itself, zero is the supremum of the heights before arrival at the hole.
These are the heights under the maintained <mechanical constraint> used in part (a). A unilateral contact with only one side of a cone would additionally require checking the sign of the <normal reaction>; the ideal constrained equations alone do not impose that condition.
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