Angle required to hit coordinates x, y, z with air ressitance

[TheShermanTanker]

The formula for the angle required for you to launch a projectile with a given velocity, gravity, distance and height difference is, taking g as gravity, v as total velocity, x as total distance on the horizontal plane and y as how high the target is above you (Negative value means the target is below you) is
tan-1((v^2 +/- square-root(v^4 - g(gx^2 + 2yv^2)))/gx). However, this assumes that the only resistive force is on the vertical plane (gravity) and that there is no horizontal resistance present (air resistance). However, the thing that I'm working on now has a hard-coded air resistance value of 1% of the object's current velocity that takes effect every 1/20 of a second (Basically imagine taking the object's velocity and multiplying it by 99% every twentieth of a second). Is there a way to factor air resistance in as well?

 

[BvU]

Hello Tanker, $\qquad$ $\qquad$ !

No analytic expression, I am afraid. So you have your numerical work (in the other thread) cut out for you

By the way, better not to call gravity force 'resisitive': it is a so-called conservative force: the work it does is converted into kinetic energy. For resisitive forces the work goes into heat and/or deformation.

 

[berkeman]

Is there a way to factor air resistance in as well?

Have you been doing any reading about "External Ballistics"? https://en.wikipedia.org/wiki/External_ballistics

I'll also page @Dr. Courtney to see if he has better links or thoughts.

 

[Dr. Courtney]

No closed for solution. As mentioned above, the needed approach is to integrate the equations of motion.

But the drag model you are working with seems unusual and slightly contrived. It might work ok for relatively short flight times.