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?
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.
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.
