$\begin{align*}\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} = \frac{1}{\sqrt{1 - (4/5)^2}} = \frac{1}{\sqrt{1 - 16/25}}= \frac{1}{\sq...$
So, the length of the meter stick in the observer's frame is

$\begin{align*}L' = \frac{1}{\gamma}L = \frac{3}{5}\mbox{ m}\end{align*}$

So, the time it takes the meter stick to pass the observer is

$\begin{align*}\frac{L'}{0.8 c} = \frac{3/5 \mbox{ m}}{(4/5) \cdot 3 \cdot 10^8} = \frac{1}{4 \cdot 10^8} \mbox{ m} = \boxed{2...$

Therefore, answer (B) is correct.

Solution to 2008 Problem 24