The trajectory of a particle $\mathbf{r}(t)$ is observed in a frame $S$ which rotates with constant angular velocity $\omega$ relative to an inertial frame $I$. Given that the time derivative in $I$ of any vector $\mathbf{u}$ is

$$\left(\frac{\mathrm{d} \mathbf{u}}{\mathrm{d} t}\right)_{I}=\dot{\mathbf{u}}+\boldsymbol{\omega} \times \mathbf{u},$$

where a dot denotes a time derivative in $S$, show that

$$m \ddot{\mathbf{r}}=\mathbf{F}-2 m \boldsymbol{\omega} \times \dot{\mathbf{r}}-m \boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r}),$$

where $\mathbf{F}$ is the force on the particle and $m$ is its mass.

Let $S$ be the frame that rotates with the Earth. Assume that the Earth is a sphere of radius $R$. Let $P$ be a point on its surface at latitude $\pi / 2-\theta$, and define vertical to be the direction normal to the Earth's surface at $P$.

(a) A particle at $P$ is released from rest in $S$ and is acted on only by gravity. Show that its initial acceleration makes an angle with the vertical of approximately

$$\frac{\omega^{2} R}{g} \sin \theta \cos \theta$$

working to lowest non-trivial order in $\omega$.

(b) Now consider a particle fired vertically upwards from $P$ with speed $v$. Assuming that terms of order $\omega^{2}$ and higher can be neglected, show that it falls back to Earth under gravity at a distance

$$\frac{4}{3} \frac{\omega v^{3}}{g^{2}} \sin \theta$$

from $P$. [You may neglect the curvature of the Earth's surface and the vertical variation of gravity.]