Answer
See answers.
Work Step by Step
From our reference frame, the muon’s internal “clock” runs slowly, allowing it to travel farther than expected. We must use the dilated time. To find the speed in the Earth’s reference frame, divide 30 km by the dilated time.
$$v=\frac{\Delta \ell_0}{\gamma \Delta t_0}=\frac{\Delta \ell_0\sqrt{1-v^2/c^2}}{\Delta t_0}$$
$$v/c=\frac{\Delta \ell_0} {\sqrt{c^2(\Delta t_0)^2}+(\Delta \ell_0)^2}$$
$$v/c=\frac{30000m} {\sqrt{(3.00\times10^8)^2(2.20\times10^{-6}s)^2+(30000m)^2}}$$
$$v/c = 0.99976$$
Calculate the kinetic energy using equation 26–5b.
$$KE = (\frac{1}{\sqrt{1-(v/c)^2}}-1)mc^2$$
$$KE = (\frac{1}{\sqrt{1-(0.9997581)^2}}-1)(105.7 MeV)=4700 MeV$$
