 Solution

Treat the atmospheric proton as stationary in the Earth frame, the incident antiproton as vertically downward, and neglect atmospheric energy loss, scattering and the Earth's curvature initially. Take $m_p{c}^2=0.9383\mathrm{GeV}$. Conservation of energy and equal outgoing energies giveEqual energies imply equal momentum magnitudes. Their transverse momenta cancel, so the two muons have equal downward components $p_z=p_{\bar{p}}/2$, whereEach muon therefore makes an angle $\theta$ with the downward vertical satisfyingThe paths are not both vertical: replacing their path lengths by $20\mathrm{km}$ would miss this two-body kinematic correction. Since $v_z=p_z{c}^2/E_{\mu }$, the proper flight time over vertical height $H$ isFor equal-energy atmospheric muon-pair survival, independent exponential particle decays with mean proper muon lifetime ${\tau }_0=2.2\mu \mathrm{s}$, both survive with probabilityEquivalently each travels about $20.55\mathrm{km}$, with proper flight time about $0.404\mu \mathrm{s}$, and survives with probability about $0.832$. The result is decay-only survival to an ideal surface, not a prediction including air interactions or detector efficiency. For a spherical Earth of radius $R_E$, the exact path length is $\ell =(R_E+H)\cos \theta -\sqrt{R_E^2-(R_E+H{)}^2{\sin }^2\theta }$; inserting $P_{\mathrm{both}}=\exp [-2\ell /({\gamma }_{\mu }{v}_{\mu }{\tau }_0)]$ gives a small curvature correction.