Q:

What angle gives the longest path for a projectile?

I know that $45°$ gives the longest range, and I feel as though $45°$ might also be the angle for the longest path of the object, but I'm not sure. How could one find what this angle is? It may also be that this angle relies on the initial velocity, so it would not be a constant.
To be clear what I mean by path, imagine that the object releases string into the air as it travels. The length of the path would be the length of that string.

A:

There is likely a neat trick to solving this, but here is a first crude stab at a solution: 
Let the projectile start from the origin at speed $v_0=1$ and angle $\theta$, tracing out the curve 
$$x(t)=\cos(\theta) t$$
$$y(t)=\sin(\theta)t-gt^2/2.$$
It will land after time $t_{max}=2\sin(\theta)/g$. The length of the trajectory is $$L=\int_0^{t_{max}} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt$$ $$=\int_0^{t_{max}} \sqrt{\cos^2(\theta)+\sin^2(\theta) + g^2t^2 - 2\sin(\theta) g t} dt$$
$$=\int_0^{t_{max}} \sqrt{1 + g^2t^2 - 2\sin(\theta) g t} dt.$$ We can reparametrize $u=gt$, $$L=(1/g)\int_0^{2\sin(\theta)} \sqrt{1 + u^2 - 2\sin(\theta) u} du.$$
We want to find $dL/d\theta=0$. By symmetry we know $\theta=\pi/2$ must be an extremum. We can either try to evaluate the integral and then find a root, or take the derivative under the integral sign (remembering that one of the boundaries also depends on $\theta$). In either case I get a messy expression in my symbolic calculator I suspect actually simplifies nicely if one massages it in the right way. 
In any case, plotting $L(\theta)$ shows that it has a maximum just below $\theta=1$ (I get an angle of $56.465^\circ$). The vertical $\theta=\pi/2$ trajectory is a local minimum: adding a bit of horizontal velocity increases the length.

