= Solution
In standard configuration the frames are inertial, their corresponding axes are parallel, S$'$ moves at $v$ along the positive $x$ axis of S, and their origins coincide at $t=t'=0$. Here $(t,x,y,z)$ and $(t',x',y',z')$ label the same event, $c$ is the <speed of light> and $\gamma=(1-v^2/c^2)^{-1/2}$.
A clock fixed in S$'$ has $\Delta x'=0$. The inverse time transformation gives
$$
\boxed{\Delta t=\gamma\Delta t'=\gamma\Delta\tau,}
$$
so the moving clock accumulates less <proper time> between its two ticks than the elapsed S-coordinate time.
Without <time dilation> the flight time from the given altitude would be $66.7\,\mathrm{\mu s}$, about $30.3$ rest-frame <mean lifetimes>, giving survival fraction only $e^{-30.3}\simeq6.8\times10^{-14}$ even at <speed> $c$. Relativistic <muons> can instead reach the ground with appreciable survival. Momentum-selected decay measurements test this clock-rate dependence experimentally.<Variation of the Rate of Decay of Mesotrons with Momentum by Bruno Rossi and David B. Hall>[Physical Review 59, 223 (1941), primary abstract: different momentum groups have different observed decay rates consistent with relativistic moving clocks]{ref}
A <mean-lifetime survival energy estimate> requires a mean flight time no greater than the dilated <mean lifetime>:
$$
\gamma\tau_0\gtrsim\frac Hc,\qquad
\boxed{\gamma\gtrsim30.3,\qquad E=\gamma m_\mu c^2\gtrsim3.21\,\mathrm{GeV}.}
$$
This corresponds to <survival probability> at least about $e^{-1}$. \b[There is no sharp minimum <energy> for an individual random decay.] For a required probability $p$, use $\exp[-H/(\gamma v\tau_0)]\ge p$. <Energy> loss in the atmosphere is neglected in this estimate. In the <muon>'s frame the atmospheric path is length-contracted to $H/\gamma\simeq660\,\mathrm m$; the ground reaches it in about $\tau_0$, giving the same <survival probability>.
For the two-body decay, use the <rest frame> of A. <Momentum> conservation makes B and the <photon> oppositely directed with equal <momentum> magnitude $p$. <Energy> conservation and the <energy–momentum relation> give
$$
m_Ac^2=E_B+pc,\qquad E_B^2=p^2c^2+m_B^2c^4.
$$
Substitution and cancellation of $E_B^2$ give
$$
\boxed{E_B=\frac{(m_A^2+m_B^2)c^2}{2m_A},\qquad
E_\gamma=\frac{(m_A^2-m_B^2)c^2}{2m_A}.}
$$
A nonzero emitted <photon> requires $m_A>m_B$.
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