Is it possible to prove that projectiles with drag will not travel in a parabola under the theory of classical mechanics? Can you prove that projectiles will travel in a parabola, provided the acceleration due to gravity $g$ does not vary significantly along the trajectory? This should take air resistance into account. I know how to prove a parabolic trajectory without air resistance (see my user page on Wikipedia, for instance, where I have successfully proven this), but with air resistance, the equations get more complicated and difficult to solve analytically.
 A: Take the equations of motion and plug in a generic parabola. It should be easy to prove that no constant coefficients of the parabola satisfy the equations of motion.
From $  \vec{a} =  \vec{g} - \beta |\vec{v}|^2 \frac{\vec{v}}{|\vec{v}|} $ by component
$$ \begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} = \begin{pmatrix} 0 \\ -g \end{pmatrix}   - \beta \sqrt{\dot{x}^2+\dot{y}^2} \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} $$
A general parabola has properties $$ \begin{pmatrix} x(t) = x_0 + v_x t \\ y(t)=y_0+v_y t - \frac{1}{2} g t^2 \end{pmatrix} $$
and when combined with the equations of motion give
$$ \begin{pmatrix} 0 \\ -g \end{pmatrix} = \begin{pmatrix} 0 \\ -g \end{pmatrix}   - \beta \sqrt{v_x^2+(v_y-g t)^2} \begin{pmatrix} v_x \\ v_y-g t \end{pmatrix} $$
which can only be true if $\beta=0$, implying that there is no air resistance.
A: Yes, you can prove that it is not a parabola. For example, see the graphics at Wikipedia.
To solve the differential equation yourself, i think the starting point would be the equation
$m\cdot \vec{a} = -k\cdot \vec{v}^2 \vec{e}_{v} - m\cdot \vec{g}$
where a is acceleration, v the speed, k some constant defining the air resistance, m the mass of your particle and g the acceleration due Earth's gravity.
A: Just solve the motion equation numerically for some set of parameters and show that those parameters do not yield a parabola. That should suffice.
A: Yes. Consider the parametric representation of a parabola, x = t, y = -t^2 (for a falling object). In the presence of drag, after a suitably long interval the equation for the object is x = C1, y = -C2 - C3 t (vertical linear motion at the terminal velocity of the object). The two equations are fundamentally different. The second is not a parabola.
