The half-life $t_{1/2}$ of a particle that decays according to
$\begin{align*}N(t) = N_0 e^{- \lambda t}\end{align*}$
is defined as the time $t$ such that $N(t) = N_0/2$ . Therefore, if half of the muons live to reach the detector, then the muons must have experienced a time of $t_{1/2}$ between their generation and their detection.

We use the time dilation formula to find the time between the generation and the detection of the muons in the laboratory frame:
$\begin{align*}t_{lab} = \frac{1}{\sqrt{1 - u^2/c^2}} t_{1/2}\end{align*}$
Therefore,
$\begin{align*}u = \frac{15 \mbox{ m}}{t_{lab}} = \frac{15 \mbox{ m} \sqrt{1 - u^2/c^2}}{ t_{1/2}}\end{align*}$
Squaring both sides and solving for $u$ gives:
$\begin{align*}u^2 = \frac{\left(15 \mbox{ m}\right)^2 c^2}{\left(15 \mbox{ m}\right)^2 + c^2 \cdot t_{1/2}^2} \Rightarrow u =...$

\section*{alternate solution:}

Using the fact that the invariant interval between the generation and the detection of the muons must have the same value in both reference frames, we have
$\begin{align*}\left(15 \mbox{ m}\right)^2 - t_{lab}^2 c^2 = - \left(2.5 \cdot 10^{-8} \mbox{ s}\right)^2 c^2 \Rightarrow t_{l...$
Therefore, the velocity in the lab frame is
$\begin{align*}\frac{15 \mbox{ m}}{t_{lab}} = \frac{15 \mbox{ m}}{\frac{\sqrt{5}}{4 \cdot 10^7} \mbox{ s}} = 12 \cdot 10^7 \sq...$
Therefore, answer (C) is correct.

Solution to 1996 Problem 48