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{\bf Question}

If a particle is projected vertically upwards to a height $h$
above a point on the ground at a northern latitude $\lambda$, show
that it strikes the ground at a point
$$\frac{4}{3}\omega\cos\lambda\sqrt{\frac{8h^3}{g}}$$ to the west
where $\omega$ is the angular velocity of the earth (Neglect air
resistance and consider only small vertical heights).

\vspace{.25in}

{\bf Answer}

Usual coordinate system - see question 2.

Newton's 2nd law: $\ds m {\bf {\ddot r}} = m {\bf g} - 2m
\mbox{\boldmath $\omega$} \times {\bf v}$

${\bf v} = \dot x {\bf i} + \dot y {\bf j} + \dot z {\bf k}
\hspace{.3in} \mbox{\boldmath $\omega$} = \omega (- \cos \lambda
{\bf i} + \sin \lambda {\bf k})$

\begin{eqnarray*} \mbox{\boldmath $\omega$} \times {\bf v} &amp; = &amp; \left|
\begin{array}{ccc} {\bf i} &amp; {\bf j} &amp; {\bf k} \\ -\omega \cos \lambda &amp;
0 &amp; \omega \sin  \lambda \\ \dot x &amp; \dot y &amp; \dot z \end{array}
\right| \\ &amp; = &amp; -\omega \sin \lambda \dot y\, {\bf i} + \omega
(\dot x \sin \lambda + \dot z \cos \lambda ) {\bf j} + (-\omega
\dot y \cos \lambda ){\bf k} \\ \\ g  &amp; = &amp; - g {\bf k}
\end{eqnarray*}

Therefore putting all this into Newton's 2nd law gives:

$\ds m( \ddot x {\bf i} + \ddot y {\bf j} + \ddot z {\bf k}) = -mg
{\bf k} - 2m\omega[ -\sin \lambda \dot y {\bf i} + (\dot x \sin
\lambda + \dot z \cos \lambda ) {\bf j} + (-\dot y \cos \lambda
){\bf k} ]$

Equating components gives:

$\begin{array}{rcl} \ddot x &amp; = &amp; 2 \omega \dot y \sin \lambda \\
\ddot y &amp; = &amp; -2 \omega( \dot x \sin \lambda + \dot z \cos
\lambda) \\ \ddot z &amp; = &amp; 2 \omega \dot y \cos \lambda - g
\end{array}$

First approximation: put $\omega = 0 \Rightarrow \ddot x = \ddot y
= 0 \hspace{.1in} \ddot z = -g$

Therefore $ x = y = 0 \hspace{.2in} z = Ut -\frac{1}{2}gt^2$ where
$U$ is the initial upward speed.

The particle reaches height $h$ when $\dot z = 0$ therefore $t =
\frac{U}{g}.$ Hence $$h = U \frac{U}{g} - \frac{1}{2} g
\frac{U^2}{g^2} = \frac{U^2}{2g}$$

The time of flight is\ \ \ $\ds t=\frac{2U}{g}.$

 Next approximation:
(insert first approximation in Coriolis terms).

$\begin{array}{rcl} \ddot x &amp; = &amp; 0 \\ \ddot y &amp; = &amp; -2\omega \cos
\lambda (U - gt) \\ \ddot z &amp; = &amp; -g \end{array}$

Thus \begin{eqnarray*} y &amp; = &amp; 2 \omega \cos \lambda  \left(
\frac{1}{6}gt^3 - \frac{1}{2}Ut^2 \right) \\  {\rm when\ }\ \  t &amp;
= &amp; 2\frac{U}{g}\ \  {\rm the\ particle\ returns\ to\ the\ ground}
\\ &amp; = &amp; 2 \omega \cos \lambda \frac{1}{6} \frac{4U^2}{g^2}\left(
g \cdot \frac{2U}{g} - 3U\right) \\ &amp; = &amp; -\frac{4}{3} \omega \cos
\lambda \frac{U^3}{g^2} \\ &amp; = &amp; -\frac{4}{3} \omega \cos \lambda
\frac{(2gh)^{\frac{3}{2}}}{g^2} \\ &amp; = &amp; -\frac{4}{3} \omega \cos
\lambda \sqrt{\frac{8h^3}{g}} \hspace{.3in}{\rm (to\ the\ west\
as\ negative)} \end{eqnarray*}



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