## College Physics (4th Edition)

The height above Earth's surface is about $6300~m$
Since three-quarters of the muons decay, the time period is two half-lives. In the muon's frame, $t = (2)(1.5~\mu s) = 3.0~\mu s$ We can find the height $h_m$ in the muon's frame: $h_m = v~t$ $h_m = (0.990)(3.0\times 10^8~m/s)(3.0\times 10^{-6}~s)$ $h_m = 891~m$ We can find the height $h_0$ in the Earth's reference: $h_m = \frac{h_0}{\gamma}$ $h_0 = h_m~\gamma$ $h_0 = \frac{h_m}{\sqrt{1-\frac{v^2}{c^2}}}$ $h_0 = \frac{891~m}{\sqrt{1-\frac{(0.990)^2}{c^2}}}$ $h_0 = \frac{891~m}{\sqrt{1-0.9801}}$ $h_0 = 6300~m$ The height above Earth's surface is about $6300~m$.