Charged pion. GONE!

Since the time taken for the pion to decay is measured in the frame of the pion, it would be the dilated time. Therefore:

$\Delta tp={ 10 }^{ -8 }s$

Expressing the 30m measured in the laboratory in terms of the speed v of the pion and the dilated time, and also in terms of the speed v of the pion and the proper time, it comes down to:

$\frac { 30 }{ \Delta tp } =\frac { v }{ \sqrt { 1-\frac { { v }^{ 2 } }{ { c }^{ 2 } } } }$

Solving the equation, you get:

$v=\sqrt { \frac { { \left[ \frac { 30 }{ \Delta tp } \right] }^{ 2 } }{ \left[ 1+{ \left[ \frac { 30 }{ \Delta tp } \right] }^{ 2 }\left[ \frac { 1 }{ { c }^{ 2 } } \right] \right] } } =2.98*{ 10 }^{ 8 }\quad (3s.f)$

Congratulations, you just earned 1 mark upon 50 in SJPO first round!

$L=30\\t^{\prime}=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{10^{-8}}{\sqrt{1-\frac{v^2}{c^2}}}\\\text{Combining the two, }\\\frac{L}{t^{\prime}}=v\\v=\frac{30\sqrt{1-\frac{v^2}{c^2}}}{10^{-18}}\\\text{Solving,}\\v\approx298000000m/s$