If anyone is curious for the answer, here goes. From part a) we get that $\Omega = 0.28$ rad/s. We can simplify the equation in the other answer (which is missing minus signs) by breaking it into vector components:

$$\begin{aligned} \frac{d\mathbf{v}}{dt} &= -\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{r})-2\mathbf\,{\mathbf{\Omega}}\times\mathbf{v}- \mu g\frac{\mathbf{v}}{v}\\ \Leftrightarrow \left<\ddot{x},\ddot{y}\right> &= \Omega^2\left<x,y\right> - 2\Omega\left<-\dot{y},\dot{x}\right> - \mu g\frac{\left<\dot{x},\dot{y}\right>}{\sqrt{\dot{x}^2+\dot{y}^2}} \end{aligned}$$

Then we use new functions $q$ = $\dot{x}$ and $p = \dot{y}$ to make 4 equations, those 2 and:

$$\begin{aligned} (3)\quad \dot{q} &= \Omega^2 x + 2\Omega p - \mu g\frac{q}{\sqrt{q^2 + p^2}}\\ (4)\quad \dot{p} &= \Omega^2 y - 2\Omega q - \mu g\frac{p}{\sqrt{q^2 + p^2}} \end{aligned} $$

so that we can solve numerically with MATLAB, which gives 0.91 seconds.

If anyone is curious for the answer, here goes. From part a) we get that $\Omega = 2.8$ rad/s. We can simplify the equation in the other answer (which is missing minus signs) by breaking it into vector components:

$$\begin{aligned} \frac{d\mathbf{v}}{dt} &= -\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{r})-2\mathbf\,{\mathbf{\Omega}}\times\mathbf{v}- \mu g\frac{\mathbf{v}}{v}\\ \Leftrightarrow \left<\ddot{x},\ddot{y}\right> &= \Omega^2\left<x,y\right> - 2\Omega\left<-\dot{y},\dot{x}\right> - \mu g\frac{\left<\dot{x},\dot{y}\right>}{\sqrt{\dot{x}^2+\dot{y}^2}} \end{aligned}$$

Then we use new functions $q$ = $\dot{x}$ and $p = \dot{y}$ to make 4 equations, those 2 and:

$$\begin{aligned} (3)\quad \dot{q} &= \Omega^2 x + 2\Omega p - \mu g\frac{q}{\sqrt{q^2 + p^2}}\\ (4)\quad \dot{p} &= \Omega^2 y - 2\Omega q - \mu g\frac{p}{\sqrt{q^2 + p^2}} \end{aligned} $$

so that we can solve numerically with MATLAB, which gives 0.91 seconds.