Projectile: $v^*w^*=gk$ for minimum launch velocity A projectile launched from $O(0,0)$ at velocity $v$ and launch angle $\theta$, passes through $P(k,h)$. The velocity of the projectile at $P$ is $w$. The slope of $OP$ is $\alpha$, i.e. $\tan\alpha=\frac hk$, and the length of $OP$ is $R$. 
$\hspace{4cm}$ 
Let $v^*$ be the minimum launch velocity for the projectile to reach $P$, and $w^* $ the corresponding minimum terminal velocity at $P$. In the course of working out  $v^*$, I noticed this neat  relationship:
$$\color{red}{\boxed{v^*w^*=gk}}\tag{1}$$
which can be proven easily using calculus. The relationship is interesting because of its symmetry and also its independence from $\theta$ and $h$. It also helps simplify the solution of projectile problems relating to minimum velocities, e.g. this question here. 

Question 1: Is it possible to derive this relationship given by $(1)$ directly without using calculus but by exploiting some geometric or kinematic symmetry?

Separately, we know that, for a projectile for a given range $R$ (on the same level), the minimum launch velocity is given by $v^*=\sqrt{gR}$. Assume that, for another given range $r$, the minimum launch velocity is $w^*=\sqrt{gr}$. Substituting in $(1)$ gives $k=\sqrt{Rr}$, i.e. $k$ is the geometric mean of $R$ and $r$. 

Question 2: Can the relationship given by $(1)$ be derived using the relationships between minimum launch velocity and range shown above, perhaps through a geometric transform of some sort?

 A: Because $w_x=v_x$ and $v_y^2=w_y^2-2gh$, the minimum speed in the origin implies you arrive with the minimum possible speed to point $(k,h)$.
I think the following picture shows a symmetry which could be useful. The minimum possible speed to reach $(k,h)$ is attained when you throw the projectile in the direction bisecting the angle between the $OP$ line and vertical axis. Because this trajectory reaches $P$ with the minimum possible speed, it also solves the "reverse" problem (by reversing the velocity direction), that is, from $P$ to reach $O$ with minimum speed. Therefore, $\vec{w}$ also bisects the angle that $OP$ makes with $\hat{y}$ at $P$, and therefore $\vec{v}\perp\vec{w}$.

By the way, proving that the maximum range in a slope (or minimum speed to a fixed range) is attained in the bisecting direction is a classical and nice problem, and it can be done without using calculus as well (here, for example).
Following with the reasoning, we then have 
$$ \begin{align}|\vec{v}\times\vec{w}|^2&=v^2w^2\qquad;\vec{v}\perp\vec{w}\\
&=(v_x w_y-w_x v_y)^2\\
&=v_x^2(v_y-w_y)^2\qquad;|v_x|=|w_x|\\
&=v_x^2(g t_f)^2\qquad;\text{where $t_f$ is time of flight}\\
&=(v_xt_f)^2g^2=k^2g^2~~.\\
\end{align}
 $$
A: This is a complement to the previously posted answer. While solving this problem I realized that the set-up also serves to prove another interesting fact. That is, the flight times of the optimal trajectories to the target depend only on $R=\sqrt{h^2+k^2}$, independent of inclination.
In the second picture below, when the minimum energy projectile thrown from $A$ reaches $D$, a "projectile" released with zero velocity from $A$ will reach point $G$. In addition, projectiles thrown with parallel velocities (including zero velocity) from the point of origin at the same time remain forming a line parallel to the original velocity at all subsequent times. Therefore, $GD$ (being the line joining the projectiles at $G$ and $D$ at time $t_f$) is parallel to $AX$ (that is, $\vec{v}\parallel AX\parallel GD$), thus, $\angle{BDG}$ is a right angle and $G$ is also in the circumference. Therefore, the flight time  for all minimum energy projectiles is the same as an object falling from $A$ to $G$, given by $\sqrt{\frac{2R}{g}}$. 

Second, more explicit derivation
Using a bit of trigonometry, we show that the smallest initial velocity $\vec{v}$ of the projectile to reach a distance of $R$ in a direction forming an angle of $\alpha$ with the horizontal is attained when the $\vec{v}$ direction bisects the angle between the vertical and $\alpha$. Part of the interest of this answer is that it does not use calculus either.
First, we calculate the range of the projectile in the direction given by $\alpha$ when being thrown with an arbitrary velocity $\vec{v}$ forming an angle $\theta$ with the horizontal. Following the same reasoning as in the previous diagram, we deduce that is $KH\parallel\vec{v}$ (in the second picture below). We can immediately see that the $\angle AHK=\theta-\alpha$, and
$$ \frac{gt_f^2}{2}= \frac{R\sin(\theta-\alpha)}{\cos\theta}~~.\tag{1}$$
 
Furthermore, it is also clear that, since there is no acceleration in the $\hat{x}$ direction,
$$ R\cos\alpha=vt_f\cos\theta~~.\tag{2}$$
From (1) and (2) we deduce 
$$R=\frac{v^2}{g}\frac{2\cos(\theta)\sin(\theta-\alpha)}{\cos^2\alpha}=\frac{v^2}{g}\frac{(\sin(2\theta-\alpha)-\sin(\alpha))}{\cos^2\alpha}~~,\tag{3}$$
from where it is very clear that the maximum range is given by $2\theta-\alpha=\pi/2~~$ (or $\theta$ bisects the angle between $AH$ and the vertical).
In turn, the maximum range is given by $\frac{v^2}{g(1+\sin\alpha)}$.
Concluding, by squaring (1)
$$\begin{align}
t_f^4&= \left(\frac{2R}{g}\right)^2\frac{\sin^2((\pi/2-\alpha)/2)}{\cos^2((\pi/2+\alpha)/2)}\qquad;\theta=(\pi/2+\alpha)/2\\
&=\left(\frac{2R}{g}\right)^2 \frac{\frac{1-\cos(\pi/2-\alpha)}{2}}{\frac{1+\cos(\pi/2+\alpha)}{2}}\\
&=\left(\frac{2R}{g}\right)^2 \frac{1-\sin\alpha}{1-\sin\alpha}\\
&=\left(\frac{2R}{g}\right)^2~~,
\end{align}
$$
therefore $t_f=\sqrt{\frac{2R}{g}}$, that is, the same result as in the first answer with $t_f$ independent of $\alpha$.
A: LATEST SOLUTION
(Using vectors)
$\hspace{5cm}$
Let $\mathbf v,\mathbf w$ be the initial (launch) and terminal velocity vectors of the projectile, and $\beta$ be the angle between them.
$\hspace{5cm}$
The projectile motion can be modelled as
(i) EITHER as a composite motion of the following:


*

*motion at constant velocity $\mathbf v$ for time $T$ (distance travelled: $vT$) and

*vertical free fall under gravity for time $T$ (distance travelled: $\frac 12gT^2$). 
(ii) OR  as a composite motion of the following:


*

*upward vertical launch at velocity $gT$ under gravity for time $T$ (distance travelled: $\frac 12gT^2$), and

*motion at constant velocity $\mathbf w$ for time $T$ (distance travelled: $wT$) and 
Let $\mathbf R$ be the direction vector for  $\vec{OP}$, and $\mathbf g$ the gravity vector. 
Distance equations for both equivalent composite motions above are as follows:
$$\begin{align}
\mathbf R&=\mathbf vT+\tfrac 12 \mathbf gT^2\tag{1}\\
\mathbf R&=\mathbf wT-\tfrac 12 \mathbf gT^2\tag{2}\\\\
\tfrac {(1)+(2)}T:\hspace {6.5cm}
(\mathbf v+\mathbf w)&=\tfrac 2T \mathbf R\tag{A}\\
\tfrac {(1)-(2)}T:\hspace{6.5cm}
(\mathbf v- \mathbf w) &= \mathbf gT\tag{B}\\\\
(\text A)\cdot (\text  B):\hspace{5cm}
(\mathbf v+\mathbf w)\cdot (\mathbf v-\mathbf w)&=2\mathbf R\cdot \mathbf g\color{lightgrey}{=2(R\sin\alpha) g}\\
v^2-w^2&=2gh\tag{I}\\\\
(\text A)\cdot (\text A):\hspace {1cm}
(\mathbf v+\mathbf w)\cdot (\mathbf v+\mathbf w)=v^2+w^2+2vw\cos\beta
&=\tfrac {4R^2}{T^2}\tag {IIa}\\
(\text B)\cdot (\text B):\hspace {1cm}
(\mathbf v-\mathbf w)\cdot (\mathbf v-\mathbf w)=v^2+w^2-2vw\cos\beta
&={g^2T^2}\tag {IIb}\\
\tfrac12\big((\text{IIa})+(\text{IIb})\big):\hspace{4.5cm} 
v^2+w^2\qquad \qquad \;\;
&=\tfrac 12 \big(\tfrac {4R^2}{T^2}+g^2T^2\big)\\
&=\tfrac 12 \underbrace{\big(\tfrac {2R}T-gT\big)^2}_{\ge 0}+2gR\\
&\ge 2gR\\
{v^*}^2+{w^*}^2&=2gR\tag{II}\\
(T^*&=\tfrac{2R}g)\\\\
\tfrac 12 \big((\text{I})\pm(\text{II})\big):\hspace{7.5cm}
{v^*}^2&=g(R+h)\\
{w^*}^2&=g(R-h)\\\\
{v^*}^2{w^*}^2&=g^2(R^2-h^2)=g^2k^2\\
\color{red}{v^*w^*}&\color{red}{=gk}
\end{align}$$

Alternatively we can bypass steps (I), (II) and arrive at the required result directly from (A), (B) as follows:
$$
\begin{align}
(\text{A})\times(\text{B}):\hspace{5cm}
(\mathbf v+\mathbf w)\times (\mathbf v - \mathbf w)
&=\tfrac 2T \mathbf R \times \mathbf gT\\
2\big|\mathbf v\times \mathbf w\big| &=2\big|\mathbf R \times \mathbf g\big|\\
vw\sin\beta&=(R\cos\alpha)g\\
&=gk\\
\color{red}{v^*w^*}&\color{red}{=gk}\hspace{1cm}
\end{align}
$$
with minimum values of $v,w$ occuring when at maximum $\sin\beta(=1)$, i.e.  at $\beta=\tfrac {\pi}2$.
