Question

A projectile is launched due north from a point in colatitude $\theta$ at an angle $\frac{\pi }{4}$ to the horizontal, and aimed at a target whose distance is y (small compared to Earth's radius R). Show that if no allowance is made for the effects of the Coriolis force, the projectile will miss its target by a distance

$x=\omega (\frac{2y^{3}}{g})^{\frac{1}{2}}(cos \theta -\frac{1}{3}sin\theta )$

Evaluate this distance if $\theta =45^{\circ}$ and y=40km. Why is it that the deviation is to the east near the north pole, but to the west both on the equator and near the south pole? (Neglect atmospheric resistance)

Answer 1

"Angle Q = pi/4 distance = y distance x = w (2y^3/g)^1/2 cos theta - 1/3 sin theta) distance theta = 45 degree & y = 40 km v = (gy/2)^1/2 + ((gv/2)^1/2 - gt) z F_coridus = 2 m v^vector times w^vector = 2m