Question

I need to find the trajectory of a projectile launched at 1000 m/s at 45 degrees to the horizontal, launched in an easterly direction from a point on the equator. I have to assume that it is a lead sphere of radius r = 5cm.

Also, what is the distance of the landing point from the launch point? Including the effects of the coriolis force and the effects of linear and quadratic air resistance. What would this trajectory look like?

As well as the trajectory of a projectile launched directly northwards?

Answer 1

The effect of the Coriolis force is an apparent deflection of the path of an object that moves within a rotating coordinate system. The object does not actually deviate from its path, but it appears to do so because of the motion of the coordinate system.

There are two reasons for this phenomenon: first, the Earth rotates eastward; and second, the tangential velocity of a point on the Earth is a function of latitude (the velocity is essentially zero at the poles and it attains a maximum value at the Equator).If the particle fired in east direction and its gain extra tangential velocity. Air Resistance directly proportional to velocity and square velocity of the particle. trajectory of the particle is parabola.

Thus, if a projectile were fired northward from a point on the Equator, the projectile would land to the east of its due north path. This variation would occur because the projectile was moving eastward faster at the Equator than was its target farther north. Similarly, if the weapon were fired toward the Equator from the North Pole, the projectile would again land to the right of its true path. In this case, the target area would have moved eastward before the shell reached it because of its greater eastward velocity. An exactly similar displacement occurs if the projectile is fired in any direction.