Find a maximizing solution to an ODE which depends on a paramater function (For the physical meaning of this problem see https://physics.stackexchange.com/questions/122818/how-should-i-throttle-my-rocket-to-reach-highest-altitude).
Given $g \in (0,\infty), k \in C^1( [0, \infty)), f \in L^1([0, \infty))$ non negative and such that $\int_0^{\infty}f \leq c < 1$, consider the Cauchy problem
$$\begin{cases}
x_f''(t)= \dfrac{f(t) - x_f'(t)^2k(x_f(t))}{1-\int_0^{\infty} f} -g,\\
x_f'(0) = 0,\\
x_f(0) = 0.
\end{cases}
$$
Find an $f$ such that for any other $\tilde f$ (which satisfy the properties above) it holds
$$ \sup_{[0,\infty)} x_{\tilde f} \leq \sup_{[0,\infty)} x_f. $$
(You may also want to consider the weak formulation with distributional derivatives).

Are there some analytic solutions to this problem for easy forms of $k$? (e.g. $k$ constant, linear, etc.)
What can we say about the regularity of solutions in the weak problem?
Can we derive from the form of $k$ some properties of the solution?
 A: I doubt you'll be satisfied with this answer. Suppose that $k(x_f) = \kappa \in (0,\infty)$ is constant (if $\kappa = 0$ everything is straight forward). Fix $C < 1$ and choose $0 < \epsilon << 1$. Set $$a = \frac{\epsilon + (1-C)g}{C}$$ and define $$f = C a \chi_{[0,\frac{1}{a}]}$$ where $\chi$ is the characteristic function of the interval $[0,\frac{1}{a}]$. Notice that $\int_0^\infty f = C$. Plugging this function into the above and considering it only over the interval $[0,\frac{1}{a}]$ we have that $$(1-C)y_f' + \kappa y_f^2 = f(t) - (1-C)g = \epsilon$$ where $y_f = x_f'$. Now, thinking about this a lot I found that $y_f = A \tanh(B t)$ is a solution to this equation with $A = \sqrt{\frac{\epsilon}{\kappa}}$ and $B = \frac{\kappa A}{1 - C}$. Thus, $x_f = \int_0^t y_f$. Note that $x_f$ satisfies all the initial conditions you've requested. Remember, this is what the solution looks like only on the interval $[0,\frac{1}{a}]$.
OK, now let's get into $x_f$ a bit more. One can show that $$ x_f(\frac{1}{a}) = \frac{A}{B} \log[\cosh(\frac{B}{a})] > \frac{1 - C}{\kappa} \log[\frac{e^{B/a}}{2}] = \sqrt{\frac{\epsilon}{\kappa}} \frac{C}{\epsilon + (1-C)g}$$ Sooo, we can take $C$ as close to 1 as we want to and $\epsilon$ really small to show that $\forall M>0$ we can find $f$ so that $$\sup_{t\in[0,\infty)} x_f > M$$ 
Maybe in your problem you don't want to assume that the mass of the rocket can be arbitrarily small? This would rule out the above construction.
