= Solution
The original diagram has a horizontal bottle neck, so weight does not enter the axial <force> balance. Gauge pressure acts on the cork's end area and supplies outward <force> $F_g=\pi a^2P_g$. The radial contact <force> is the contact pressure times the cylindrical contact area, $N=2\pi a(l-x)P_c$. The limiting friction is $\mu N$, so outward sliding occurs when
$$
\pi a^2P_g>2\pi a(l-x)\mu P_c,\qquad \boxed{P_g>\frac{2(l-x)\mu P_c}{a}.}
$$
The assumption of equal static and <kinetic friction> coefficients removes a separate <force> drop when slipping starts.
Initially the pressure is marginal at $x=0$, so $\pi a^2P_g=2\pi al\mu P_c$. During extraction the contact length decreases, while the pressure <force> stays constant. The net <force> is therefore $F(x)=\pi a^2P_gx/l$. The cork's <mass> is $m_c=\rho\pi a^2l$, giving the <pressure-driven cork ejection> equation
$$
\boxed{\ddot x=b^2x,\qquad b^2=\frac{P_g}{\rho l^2}.}
$$
For $x(0)=0$ and $\dot x(0)=v_0>0$, solving the constant-coefficient equation gives
$$
\boxed{x(t)=c\sinh(bt),\qquad b=\frac1l\sqrt{\frac{P_g}{\rho}},\qquad c=\frac{v_0}{b}=v_0l\sqrt{\frac\rho{P_g}}.}
$$
The equilibrium is unstable: as the cork moves out, friction decreases and the unbalanced <force> grows.
Multiplying the equation of motion by $\dot x$ and integrating gives $\dot x^2=v_0^2+b^2x^2$. The contact ends at $x=l$, so
$$
\boxed{v_{\mathrm{exit}}^2=v_0^2+\frac{P_g}{\rho},\qquad v_{\mathrm{exit}}\longrightarrow\sqrt{\frac{P_g}{\rho}}\quad(v_0\to0^+).}
$$
Equivalently, the net <work> is $\int_0^lF(x)dx=\pi a^2P_gl/2$. Exactly zero initial <velocity> would leave the ideal marginal configuration stationary. The exit time $b^{-1}\operatorname{arsinh}(bl/v_0)$ diverges as $v_0\to0^+$, even though the limiting exit <speed> is finite.
The density is $500\,\mathrm{kg\,m^{-3}}$, so $v_{\mathrm{exit}}\simeq\sqrt{400000/500}=28.3\,\mathrm{m\,s^{-1}}$. For an ideal upward launch at this <speed>, <conservation of mechanical energy> gives height above the exit
$$
\boxed{h=\frac{v_{\mathrm{exit}}^2}{2g}=\frac{P_g}{2\rho g}\simeq40.8\,\mathrm m\approx41\,\mathrm m.}
$$
This neglects air resistance and the small gravitational <work> during extraction compared with the launch <energy>. Actual gas pressure need not stay constant, so it is an idealized estimate.
Adding a stopper that doubles the moving <mass>, while preserving the same pressured area and contact friction, leaves the extraction <work> unchanged. The squared exit <speed> and the ballistic height are therefore halved:
$$
\boxed{v_{\mathrm{stopper}}=\frac{v_{\mathrm{exit}}}{\sqrt2},\qquad h_{\mathrm{stopper}}=\frac h2\simeq20.4\,\mathrm m.}
$$
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