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For a cone of half angle $\pi /4$ draining through a small circular aperture of radius $ϵ$, the horizontal area at height $h$ is $\pi h^2$. Torricelli's law gives speed $\sqrt{2gh}$, so conservation of mass gives $h^2\dot{h}=-{ϵ}^2\sqrt{2gh}$. Integration yields the displayed leading drainage time when $h_0\gg ϵ$. The final layer $h=O(ϵ)$ is outside the quasi-steady approximation and contributes only order $\sqrt{ϵ/g}$, compared with the leading time of order $h_0^{5/2}/({ϵ}^2\sqrt{g})$. This is an ideal inviscid estimate without a discharge coefficient.

For constant fluid density $\rho$, steady Euler equations for an inviscid fluid under a conservative force per unit mass $-\nabla \Phi$ read $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho -\nabla \Phi$. Here $\mathbf{u}$ is velocity, $p$ pressure and $\Phi$ potential per unit mass. Dotting with $\mathbf{u}$ givesThis is the Bernoulli equation. The constants can differ between streamlines. If density varies, the same derivation for a barotropic fluid replaces $p/\rho$ by ${\int }^{p}d{p}^{\prime }/\rho (p^{\prime })$; constant density is the appropriate water approximation here.

For the draining cone, the radius at height $h$ is $h$ and the horizontal area is $\pi h^2$. Both the free surface and the jet are at atmospheric pressure; their gravitational potential difference is approximately $gh$. The speed of the descending surface is smaller than the jet speed by the area ratio ${ϵ}^2/h^2$, so it can be neglected when $h\gg ϵ$. The Bernoulli equation then gives Torricelli's law, $U\simeq \sqrt{2gh}$. Conservation of mass givesHence the leading drainage time of a conical tank isRemoving the tip places the aperture at height of order $ϵ$, and the quasi-steady calculation fails in the last layer of that depth. Integrating down only to $h=O(ϵ)$ changes the displayed leading time by relative order $(ϵ/h_0{)}^{5/2}$; replacing the actual head $h-ϵ$ by $h$ introduces a larger, but still vanishing, relative correction of order $ϵ/h_0$. The formula is the ideal inviscid estimate under the question's stated approximation, not an exact law for the final transient or a real contracted jet.

If $z(t)$ is the water depth and $A(z)=\pi r(z{)}^2$ is the horizontal cross-sectional area, Torricelli's law gives the volume flux through the hole asConservation of volume givesFor a prescribed constant fall rate $\dot{z}=-\alpha$, $\alpha >0$, this requiresConsequentlyso the container radius must be proportional to the fourth root of height.