Answer
In the Earth rest frame, the pion decays at an altitude of $~~110~km$
Work Step by Step
We can find $\gamma$:
$E = \gamma~mc^2$
$\gamma = \frac{E}{mc^2}$
$\gamma = \frac{1.35\times 10^5~MeV}{139.6~MeV}$
$\gamma = 967$
We can find $\Delta t$ in the Earth frame:
$\Delta t = \gamma~\Delta t_0$
$\Delta t = (967)(35.0~ns)$
$\Delta t = 33.845~\mu s$
We can find the pion's speed:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$1-\beta^2 = \frac{1}{\gamma^2}$
$\beta^2 = 1-\frac{1}{\gamma^2}$
$\beta = \sqrt{1-\frac{1}{\gamma^2}}$
$\beta = \sqrt{1-\frac{1}{(967)^2}}$
$\beta = 0.999999465$
We can find the distance the pion travels before it decays:
$d = (0.999999465~c)(33.845~\mu s)$
$d = (0.999999465)~(3.0\times 10^8~m/s)(33.845~\mu s)$
$d = 10~km$
In the Earth rest frame, the pion decays at an altitude of $~~110~km$
