Curves & Motion · Challenge problem

A ball's position (metres, seconds) is $\mathbf r(t) = \left(10t,\ 20t - 5t^{2}\right)$ for $0 \le t \le 4$. At the launch the radius of curvature of its path is greater than $80$ m, and at the top it is only $10$ m. At what time $t$ is the radius of curvature first equal to $80$ m? Give an exact answer.