Maximizing Time of Flight in Projectile Motion Is (or How is) it possible to maximize the time of flight of projectile subject to the following conditions?
Given :


*

*Fixed horizontal range 

*Interval in which velocity lies
For example, let the range be 100 feet. The velocity of projectile lies in the closed interval [40√2,80]$\frac{ft}{s}$. Given acceleration due to gravity is 32$\frac{ft}{s^2}$. Then, ignoring frictional forces (more precisely, viscous drag) what will be the maximum value of time of flight of projectile?
 A: The range is given by
$$R(v_0,\theta_0)=\frac{v_0^2\sin2\theta_0}g$$
The flight time is given by:
$$\tau(v_0,\theta_0)=\frac{2v_0\sin\theta_0}g$$
Maximum point is given by:
$$\nabla\tau(v_0,\theta_0)=0$$
Our constrain here is that $R$ is fixed. so $\lambda R$ is also fixed. So adding it to $\tau(v_0,\theta_0)$ doesn't alter $\nabla\tau(v_0,\theta_0)$.
Lagrange has invented this trick. That $\lambda$ is the equalizing constant which corrects the units, scaling, etc. Now let's use is.
$$\nabla(\tau(v_0,\theta_0)+\lambda R)=0$$
$$\nabla(\frac{2v_0\sin\theta_0}g+\lambda\frac{v_0^2\sin2\theta_0}g)=0$$
$$\nabla(2v_0\sin\theta_0+\lambda v_0^2\sin2\theta_0)=0$$
$$(\sin\theta_0+\lambda v_0 \sin2\theta_0,\cos\theta_0+\lambda v_0\cos2\theta_0)=0$$
Gives,
$$-\lambda v_0=\frac1{2\cos\theta_0}\ \ ,\ \ \cos\theta_0=-\lambda v_0\cos2\theta_0$$
$$\sin^2\theta_0+\cos^2\theta_0=1\ \ , \ \ v_0=-\frac1{2\lambda\cos\theta_0}$$
Considering $L_- \le v_0 \le L_+$, we get,
$$-\frac1{2\lambda L_+}\le \cos\theta_0 \le -\frac1{2\lambda L_-}$$
So $\theta_0$ can be anything as long as it satisfies the condition above, and in the upper line, we have $v_0$ as a function of $\theta_0$. So everything is solved.
Except one thing, I haven't studied how to find $\lambda$ yet. :D So sorry. Others familiar with Lagrange's method may help me complete my answer.
A: Use the trajectory equation for uniform acceleration and take the derivative set to zero to find critical points. Use maximum. Usually theta is 45 ignoring draft. 
