5(b)

The limiting velocity is -mg/k. The velocity is 80% of this when

$$-.8 \frac{mg}{k} = - \frac{mg}{k}\left(1 - {\rm e}^{-kt/m}\right)$$

so $$\exp(-kt/m) = .2$$ and $$t = \frac{m}{k} \ln 5$$.