Finding the Angle in Projectile Motion

[Robotics 1764]

Homework Statement

Hello, I am attempting to derive an equation for some pretty specific needs. The equation I need is one that solves for the angle needed to launch a projectile at a given velocity to an object at a given distance and height different from launcher. The catch is that I would like to include the acceleration due to air resistance (which I can find later and input) as an acceleration in the x direction.

The goal is to solve for theta (I will use an o)

Homework Equations

The only variables that can be present in the equation are:
ax, ay (gravity), x, y, v (velocity launched at, actually, it is the speed, I don't know the angle)

The equations I have been attempting to use are:

y = vy*t + (g/2)*t^2
x = vx*t + (a/2)*t^2
v = sqrt(vx^2 + vy^2)
sin o = vy/v
cos o = vx/v

The Attempt at a Solution

I keep trying to derive but keep ending up circling (deriving until I accidentally reach a variable I tried to get rid of earlier). Does anyone have an equation already derived they could show me that they could also explain how they derived, or know how I could go about deriving my goal equation successfully? Or any other formulae I should be using?

Thanks.

 

[tiny-tim]

Welcome to PF!

The equation I need is one that solves for the angle needed to launch a projectile at a given velocity to an object at a given distance and height different from launcher.

The catch is that I would like to include the acceleration due to air resistance (which I can find later and input) as an acceleration in the x direction.
…
The equations I have been attempting to use are:

y = vy*t + (g/2)*t^2
x = vx*t + (a/2)*t^2
v = sqrt(vx^2 + vy^2)
sin o = vy/v
cos o = vx/v

Hi Robotics 1764! Welcome to PF!

(air resistance would never be like that, but anyway …)

Hint: since you have constant acceleration in both the x and y directions, try combining them (accelerations obey the vector law of addition, of course) into a single acceleration, and then use coordinates parallel and perpendicular to that direction.