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{\bf Question}

If a projectile is fired due east from a point on the surface of
the earth at a northern latitude $\lambda$ with a speed $V_0$ and
at an angle of inclination to the horizontal of $\alpha$, show
that the lateral deflection when the particle strikes the earth is
$$d = \frac{4V_0^3}{g} \omega\sin\lambda\sin^2\alpha\cos\alpha.$$

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{\bf Answer}

Same set up as 2b.

Initial conditions: $\ds \dot y = V_0 \cos \alpha, \dot z = V_0
\sin \alpha;\ \ \   x = y = z = 0.$

First approximation: set $\omega = 0$

$\ds y = (V_0 \cos \alpha)t$

$\ds z = (V_0 \sin \alpha)t - \frac{1}{2}gt^2$

Returns to ground when $\ds t = \frac{2V_0 \sin \alpha}{g} (*)$

Next approximation:

$\left.\begin{array}{rcl} \ddot x & = & 2 \omega \sin \lambda V_0
\cos \alpha \\ \ddot y & = & -2 \omega \cos \lambda (V_0 \sin
\alpha - g t) \\  \ddot z & = & 2 \omega \cos \lambda V_0 \cos
\alpha - g \end{array}\right\}$ From N2 as in 2b.

Thus $x = V_0 \omega \sin \lambda \cos \alpha t^2$

Putting this in (*) gives $\ds x = V_0 \omega \sin \lambda \cos
\alpha \frac{4v_0^2 \sin^2 \alpha}{g^2}$ (as required)

i.e. projectile deflected south by this amount.


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