Question

you answer Angular velocity of the Earth: ω 7.27×10-5 rad/s Problem 1 (10 points): In problem 6 of the homework you found the southerly deflection of a projectile fired eastward to be 92 where α is the launch angle. Calculate this deflection for a projectile that reaches a height of 100. m with a launch angle 45°, at a) the equator. b) Miami (latitude 25°) c) a location near the north pole. Why can't we be exactly at the north pole to do this calculation?

Answer 1

The projectile height is

$h=\frac{v_0^2\sin^2\alpha}{2g}$

Given the launch angle is 45 degree and height is 100 m

Hence,

$100=\frac{v_0^2}{4g}=\frac{v_0^2}{4\times 9.8}$

Therefore the launch velocity is

$v_0=\sqrt{4\times 9.8\times 100}=77.201\,m/s$

a) The deflection at equator

At equator $\lambda = 0$ Hence $\sin\lambda = 0$

Therefore deflection is zero.

$x=0$

b) At Miami $\lambda = 25^o$

$x=\frac{4\omega v_0^3}{g^2}\sin\lambda \cos\alpha \sin^2\alpha=\frac{4\times 7.27\times 10^{-5}\times 77.201^3}{9.8^2}\sin 25^o\cos 45^o\sin^245^o$

This gives

$x=0.2082\,\,m$

c) At a location near the north pole

the latitude is almost 90 degrees

$\lambda = 90^o$

therefore the deflection is

$x=\frac{4\times 7.27\times 10^{-5}\times 77.201^3}{9.8^2}\sin 90^o\cos 45^o\sin^245^o=0.493\,\,m$

The point where the axis of rotation meets the earth's surface wobbles slightly. And that's why the latitude angle in the calculation is not exactly 90 degrees at the exact north pole. This is the reason you cannot do this calculation at the exact north pole