Projectile angle and height for long-distance flight across Earth's curvature

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if a body is projected up with some velocity (not much)& at an angle (not equal to 90),such that it strikes back the earth,but at a place far far away from initial point
say from India to america.


can anyone tell me the procedure to calculate the angle which it will make with the ground where it will strike and the maximum height reached by it with respect to ground vertically bellow it at that instant?



assuming Earth to be a perfect sphere and taking suitable variables for angle of projection, initial velocity mass of Earth & body & radius of Earth .(neglect air resistance).
 
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I saw a very similar problem in one of the Physics forums just a couple of days back. Search for it using the keyword "orbit".
 
The question does not make any sense. Firing a projectile from India to america with some velocity. On the surface intuition senses that a 45% angle gives the longest range. But if the projectile is rotating the spin causes some lift and an angle just above 45% gives the longest range. You also have to take into account air density, temperature and windage.
 
You also have to take into account the spin of the earth, the latitudes of the points of launch and impact, the variation of the density of air with altitude, just to name a few. That does not make the question senseless. All this is calculated for launching ICBMs.

We can try our hand at the simplest version. On a non-rotating spherically symmetric airless earth, a missile is launched in the equatorial plane. Where will the point of impact be, given the initial velocity?