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+### Statement
+
+$14.1.2.$ [Insert the problem statement]
+
+### Solution
+
+### Problem
+
+$14.1.2.$ After \(10^{-8}\) seconds following the passage of a \(\pi^0\)-meson through detector A, both detectors A and B registered gamma (\(\gamma\)) quanta, which were produced from the decay of the \(\pi^0\)-meson:
+\[
+\pi^0 \rightarrow \gamma + \gamma
+\]
+The distance between the detectors is 1 meter. What was the speed of the \(\pi^0\)-meson?
+
+![|1530x294, 100%](../../img/14.1.2/14.1.2.png)
+
+### Solution
+
+Since the detectors simultaneously registered the gamma quanta, this means the quanta traveled equal distances (their speeds are equal and equal to \(c\), the speed of light). Therefore, the decay point of the \(\pi^0\)-meson is in the middle between the detectors: \(\frac{\ell}{2}\).
+
+According to the problem, the total time from the meson's flight to the decay point and then the flight of the gamma quantum to the detector is \(t = 10^{-8}\) s:
+\[
+\frac{\ell}{2} \left( \frac{1}{\upsilon} + \frac{1}{c} \right) = t
+\]
+Here, \(\upsilon\) is the unknown speed of the meson. Solving this equation for \(\upsilon\):
+\[
+\upsilon = \frac{\ell c}{2ct - \ell}
+\]
+Substituting the numerical values into the expression gives the speed of the \(\pi^0\)-meson:
+\[
+\upsilon = 6 \cdot 10^{7} \text{ m/s}
+\]
+
+### Answer
+
+\[
+\upsilon = 6 \cdot 10^{7} \text{ m/s}
+\]
+
+#### Answer
+
+[Insert a concise answer or boxed result]