 Solution

In terms of the positive cylindrical radius $r=-z$, the conserved mechanical energy becomesThe effective potential $U$ tends to infinity at both ends of $(0,\infty )$ for $v>0$, and $U^{\prime \prime }(r)=3a^2{v}^2/r^4>0$. Thus its two turning points enclose the allowed radial motion. At a turning point, $\dot{r}=0$, soBesides $r=a$, the positive root of a polynomial isthe other root of a polynomial is negative. The quadratic polynomial $Q(r)=2g{r}^2-v^{2}r-a{v}^2$ is negative at zero and has just one positive root of a polynomial. Since $Q(a)=2a(ga-v^2)$, we have $r_{\ast }<a$ when $v^2<ga$, and $r_{\ast }>a$ when $v^2>ga$.

Height is $z=-r$, so its maximum occurs at the smaller positive turning point. HenceIndeed, the initial vertical acceleration isBelow the critical speed the particle first rises to $z=-r_{\ast }$; 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.