Can a glider maintain a constant forward velocity component? I have a puzzle which asks me to consider a (engineless) glider descending at a constant rate and flying at a constant forward velocity (in still air). The question asks how to derive sink rate, $v_s = \frac {Av^{-2}} {mg} $(vertical velocity), assuming that the drag acts horizontally (because the descent angle is small).
Question 1:
My best stab at a  force diagram (below) indicates that a constant forward speed is impossible! By the way I'm only considering the drag in order to produce lift (induced drag).

How could forward velocity be constant? Is it a reasonable approximation? Are there any other forces on the glider?
Question 2 - Trying to derive sink rate,$v_s$
Leaving that aside and assuming speed v is constant, I can see drag, $D = Av^{-2}$ is the rate of energy loss (with a small $v_s$ so that $v \approx \frac {dx} {dt}$ where x is position) because
$$ Power, P = \frac{d} {dt} \int{D dx} = \frac{d} {dt} \int{Av^{-2} dx} = Av^{-2} \frac {dx} {dt} \approx Av^{-1}$$
The power developed by the weight of the object is the rate of change of G.P.E. $ = mgv_s$
$$\Rightarrow mgv_s \approx Av^{-1} \Rightarrow v_s \approx \frac {Av^{-1}} {mg}$$
Where have I gone wrong?
 A: As the comments say, you need some horizontal component of lift to overcome drag. And your drag is more than only the induced component - the friction losses are of the same magnitude.

Forces acting on glider (own work)
Note that I shifted the vectors such that it can be shown that the force triangle is closed, and the forces are in equilibrium. $v_{\infty}$ is the speed of air relative to the glider caused by the glider's motion. The inclination is needed to tilt the lift vector (which per definition is orthogonal to the airspeed vector) forward, resulting in a horizontal component which equals drag.
For a very crude first estimate of minimum sink speed of a glider, just take 100 and subtract the best glide ratio. The result is very close to its minimum sink expressed as centimeters per second. Works only for gliders!
On a more serious note, in order to get to a useful solution for the sink speed $v_z$, you start with potential energy loss over time:
$$\frac{dE_{pot}}{dt} = m\cdot g \cdot\frac{dh}{dt} \approx W\cdot v$$
With the equilibrium of forces ($L$ = lift, $D$ = drag)
$$L = -m\cdot g\cdot cos\gamma \;\text{and}\; D = -L\cdot tan\gamma $$
you can write
$$m\cdot g \cdot\frac{dh}{dt} = m\cdot g \cdot cos\gamma\cdot v\cdot tan\gamma \;\text{and}\;  tan\gamma = -\frac{D}{L} = -\frac{c_D}{c_L}$$
$$v_z = -v\cdot sin\gamma = v\cdot\frac{c_D}{\sqrt{c_L^2+c_D^2}} \approx v\cdot\frac{c_D}{c_L}$$
Next you need some approximation for $c_D$:
$$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$
so you can write
$$v_z \approx v\cdot\left(\frac{c_{D0}}{c_L}+\frac{c_L}{\pi\cdot AR \cdot \epsilon} \right)$$
Nomenclature:
$g\;\;\;\;\;\;$Gravitational acceleration
$v\;\;\;\;\;\;$Flight speed
$c_{D}\;\;\;\;$Drag coefficient
$c_{D0}\;\;\;$Zero lift drag coefficient
$c_{L}\;\;\;\;$Lift coefficient 
$\gamma\;\;\;\;\;\;$Flight path angle, positive when pointing up from the horizontal
$\pi\;\;\;\;\;\;$3.14159…
$AR\;\;\;$Wing aspect ratio (span squared over area)
$\epsilon\;\;\;\;\;\;\,$Oswald factor, normally between 0.7 and 1. For gliders 0.98 is typical
Yes, sink speed is positive downwards.
A: @PeterKämpf is right. Let me just try what might be a simpler way to think about it.
The plane has a certain speed $v$ and a certain drag $D$, so the energy loss due to drag is $vD$.
The plane has a certain weight $mg$, and is descending at sink rate $v_s$, so it is changing potential energy into kinetic energy at a rate $mgv_s$.
These two energies have to be equal. It's as simple as that.
By the way, at a given speed, induced drag $D$ is just a certain fraction of lift force, so it is a certain fraction of $mg$. Most good gliders have a glide ratio of about 30:1, so the drag should be about $mg/30$, and that includes parasitic as well as induced.
A: 
As shown, $AB$ is the wing of the glider.  The yellow arrow is the speed direction $V$ of the glider.  $G$ is the weight of the glider.  $G$ has two components, one is a component $G_t$ parallel to the velocity direction, and one is a component $G_n$ perpendicular to the velocity direction.  There are also two components of the air force applied to the wing, one is the component $D$ parallel to the speed direction, and the other is the component $L$ perpendicular to the speed direction.
If the speed $V$ of the glider is low, then $G_t> D$ and $G_n> L$.  Thus, under the thrust of $G_t$, the speed $V$ of the glider increases.  As the speed $V$ increases, $D$ will increase, and $L$ will increase. Finally, $G_t = D$, $L = G_n$, and the glider will glide along the yellow line at a uniform speed.
From this analysis of the glider, we can see that the flight of the glider is a thrust flight.  Without gravity as the thrust, the glider cannot fly and cannot generate lift.

Why does an object move down an inclined plane? Because the component of gravity is parallel to the inclined plane. It pushes the object down an inclined plane. Can't "lift" push an object down an inclined plane here? It can't, because lift is perpendicular to the inclined plane, and it has no component in the parallel direction of the inclined plane. Here, gravity is the force that pushes an object down an inclined plane. Because gravity has components in the direction parallel to the inclined plane. The same is true of gliders. The glider is also driven downward by gravity.
