State what the vectors $\mathbf{a}, \mathbf{r}, \mathbf{v}$ and $\boldsymbol{\omega}$ represent in the following equation:

$$\mathbf{a}=\mathbf{g}-2 \boldsymbol{\omega} \times \mathbf{v}-\boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r})$$

where $\mathbf{g}$ is the acceleration due to gravity.

Assume that the radius of the Earth is $6 \times 10^{6} \mathrm{~m}$, that $|\mathrm{g}|=10 \mathrm{~ms}^{-2}$, and that there are $9 \times 10^{4}$ seconds in a day. Use these data to determine roughly the order of magnitude of each term on the right hand side of $(*)$ in the case of a particle dropped from a point at height $20 \mathrm{~m}$ above the surface of the Earth.

Taking again $|\mathbf{g}|=10 \mathrm{~ms}^{-2}$, find the time $T$ of the particle's fall in the absence of rotation.

Use a suitable approximation scheme to show that

$$\mathbf{R} \approx \mathbf{R}_{0}-\frac{1}{3} \boldsymbol{\omega} \times \mathbf{g} T^{3}-\frac{1}{2} \boldsymbol{\omega} \times\left(\boldsymbol{\omega} \times \mathbf{R}_{0}\right) T^{2},$$

where $\mathbf{R}$ is the position vector of the point at which the particle lands, and $\mathbf{R}_{0}$ is the position vector of the point at which the particle would have landed in the absence of rotation.

The particle is dropped at latitude $45^{\circ}$. Find expressions for the approximate northerly and easterly displacements of $\mathbf{R}$ from $\mathbf{R}_{0}$ in terms of $\omega, g, R_{0}$ (the magnitudes of $\boldsymbol{\omega}, \mathbf{g}$ and $\mathbf{R}_{0}$, respectively), and $T$. You should ignore the curvature of the Earth's surface.