How can I find the maximum height if I have changing acceleration? I'd like to find the maximum height that a projectile reaches in motion if I have a non-constant acceleration. 
How would I do this? What variables do I need to know in order to calculate this?
Thank you!
 A: If the acceleration is non-constant, then its depends on a number of parameters. 
If it depends from it's position
$$ \vec a = \vec a(\vec r) $$
It doesn't really matter what these parameter's are or how many.
On this projectile the forces that are acted are it's weight and another force $\vec f$. According to Newton's third law 
$$ \sum \vec F = m \vec a $$
So $$ \sum \vec F = m \vec a(\vec r)= m \vec g + \vec f $$ 
What you have to make use of is the Kinetic energy as well as the potential in order to find the maximum height.
$$ W = \int \sum \vec F \cdot d \vec r = \int ( \vec W + \vec f) \cdot d \vec r$$ 
From the work-energy theorem:
$$ K_f - K_i = V_i - V_f + \int \vec f \cdot d\vec r$$
When the projectile has its maximum height then it has no velocity. Therefore
$$ h_{max} = \frac{ K_i + m \int ( \vec a(\vec r)- \vec g) \cdot d \vec r}{mg} $$
If it depents on time
If $ \vec a = \vec a(t) $ then 
$$ \vec v - \vec v_0 = \int_{t_0}^t \vec a(t)dt \Rightarrow \vec v = \vec v_0 + \int_{t_0}^t \vec a(t) dt $$
$$ \vec h-\vec h_0 = \vec v_0t + \int_{t_0}^t ( \int_{t_0}^t \vec a(t) dt ) dt $$$$\vec h = \vec h_0 + \vec v_0t + \int_{t_0}^t ( \int_{t_0}^t \vec a(t) dt ) dt$$
So you find when the velocity becomes zero and then you find the height for this time. It is possible for the velocity to become zero in more than one times and the heights not to be the same. Sto the maximum height of the motion is the biggest of these heights.
The truth is that the acceleration might be dependent on both position and time or more parameters. To solve these problems you need to know more mathematical technics which i am not confident enough to present at the moment. I'd love to see another answer or when I am confident about the answer i will post it.
A: At the maximum height, the vertical velocity is zero. So if you have an expression for the vertical velocity, just set it equal to zero and solve for vertical position.
