The apex of parabolic motion forms an ellipse of constant ellipticity. I am not sure how well-known this is idea is, but here is a .gif illustrating it:

Basically, the set of highest points of parabolic motion at constant initial velocity forms an ellipse, with eccentricity which is independent of both the initial velocity and gravitational acceleration. It's pretty easy to see that it's true, and I will work it out  here for completeness. The highest point of the ellipse (which will be the semi-minor axis $b$) is
$h=2b=\frac{v^2}{2g}$
where $v$ is the initial velocity and $g$ is the gravitational acceleration. The semi-major axis will be the largest horizontal distance any path makes before it peaks, the time of which we can find with kinematic equations assuming the y-velocity goes to zero
$v_y(t)=0=v\sin\theta -gt\rightarrow t=\frac{v}{g}\sin\theta$
Plug this into the kinematic equation for distance and minimize:
$x(t)=v\cos\theta t=\frac{v^2}{g}\sin\theta\cos\theta$
This maximum occurs for $\theta=\pi/4$, so $a=\frac{v^2}{2g}$ and the eccentricity is $e=\sqrt{3}/2$, and does not depend on the initial velocity or the gravitational acceleration.
Now, I think that's pretty amazing; at this point I've convinced myself there is no ellagant ("simple") way of seeing this fact, but I am interested if it might be a special case of something else. 
For instance, my thought is that it is somehow related to the lengths of geodesics on a space of positive curvature, since we have curves from the origin of motion to critical points in a field of constant acceleration. Of course, it can't quite be that because the space actually seems to be more like a cylinder than a sphere, and I'm not sure what to do about the axes; I guess they would have to be measured with respect to a 3-space which my cylindrical surface is embedded.
Anyway; does anyone know anything about this problem, or if it represents a specific case of some more interesting geometrical result?
EDIT: Well, there has been very little interest in this question, and after ~5 years I am still fascinated by it. So what else can be said which might evoke some interest? 
This is 2D kinematic motion, which means each of the two directions of motion are described by a polynomial. The specific polynomials here are
$$(x(t),y(t))=(v_{x0}t+x_0,-1/2gt^2+v_{y,0}t+y_0)$$
Of course, the answer did not depend on any of the parameters of this problem ($v_{x0},v_{y0},g,x_0,y_0$), but perhaps it depends on the order of the polynomials...at least, that's the only choice I'm left with.
So, perhaps each particular combination of polynomial order $(n,m)$ (where $(n,m)=(1,2)$ for this case) results in a different value of the eccentricity?
Does anyone know if it is possible to model ellipses with polynomials?
 A: Here's a way to generalize the geometry of the situation to non-parabola conics; unfortunately, the locus of vertices is not itself a conic (except in the parabolic case).

First, I'll reiterate OP's setup this way: 

Consider the path of a projectile, subject to gravity ($g$), that leaves the origin with a speed $v$ at an angle $\theta$ with the horizontal. Parameterized by time ($t$) the path is given by
  $$(x,y) = \left( v t \cos\theta, -\tfrac12 gt^2 + vt \sin\theta\right) \tag{1}$$

Eliminating $t$, we find that the projectile's parabolic path has this Cartesian equation
$$2 y v^2 \cos^2\theta = - g x^2 + 2 x v^2 \cos\theta \sin\theta \tag{2}$$
or, in standard form, where I'll take the opportunity to define $p := v^2/g$:

$$\left( y - \tfrac{1}{2}p\sin^2\theta \right) = - \frac{1}{2p\cos^2\theta} \left( x - \tfrac12 p \sin2\theta\right)^2 \tag{3}$$

Noting that $\sin^2\theta = \tfrac12(1-\cos2\theta)$, we see that the projectile's vertex, $(\overline{x},\overline{y})$, satisfies
$$\left( \frac{2}{p} \overline{x} \right)^2 + \left(1 - \frac{4}{p} \overline{y}\right)^2 = \sin^22\theta + \cos^22\theta = 1 \quad\to\quad \frac{4}{p^2} \overline{x}^2 + \frac{16}{p^2}\left(\overline{y}-\tfrac14p\right)^2 = 1 \tag{4}$$
which describes an ellipse of eccentricity $e = \sqrt{3}/2$, independent of $p$ (and, thus, parameters $g$ and $v$). This reconfirms OP's observation.

To generalize, I'll ignore the physics and concentrate on the geometry. We can capture most of the original setup by considering a vertically-oriented conic that passes through the origin, whose tangent line there makes an angle $\theta$ with the $x$-axis. This is not enough information to determine the conic, and it also does not account for the fact that OP's parabolas get narrower as they get steeper; it's not entirely clear which additional assumption is appropriate to fill this gap. This approach looks to equation $(3)$, recalling that the reciprocal of the coefficient of the squared term is the latus rectum of the parabola; accordingly, we'll take our $\theta$-conic to have semi-latus-rectum $2\cos^2\theta$.
To apply these conditions, consider that the general equation of a vertically-oriented non-parabola conic with eccentricity $e$, major radius $a$, and center $(h,k)$ has equation
$$(x-h)^2 + (1-e^2) (y-k)^2 = a^2 (1-e^2) \tag{5}$$ 
Then, we have the following:
$$\begin{align}
h^2 + (1-e^2) k^2 &= \phantom{-}a^2 ( 1 - e^2 ) & \text{(contains origin)} \tag{6a} \\
h \cos\theta &= -k (1-e^2) \sin\theta &\text{(tangent angle $\theta$ at origin)} \tag{6b} \\
\pm a(1-e^2) &= \phantom{-}p \cos^2 \theta &\text{(semi-latus-rectum length $p\cos^2\theta$)} \tag{6c}
\end{align}$$
where the "$\pm$" in $(6c)$ is "$+$" for ellipses and "$-$" for hyperbolas. We can solve this system for $a$, $h$, $k$ to get
$$a = \pm \frac{p\cos^2\theta}{1 - e^2} \qquad h = \pm \frac{p \sin\theta \cos^2\theta}{
 \sqrt{1 - e^2 \sin^2\theta}} \qquad
k = \mp \frac{p \cos^3\theta}{(1-e^2)\sqrt{1 - e^2 \sin^2\theta}} \tag{7}$$
(A second solutions switches the signs on $h$ and $k$. This corresponds to an identical conic situated on the other side of the tangent line.) Substituting into $(5)$, we have
$$\left(\,x^2 + (1 - e^2) y^2\,\right) \sqrt{1 - e^2 \sin^2\theta} \;=\;\pm 
 2 p \cos^2\theta \left(\,x \sin\theta - y \cos\theta\,\right)\tag{8}$$
When $e=1$, taking "$\pm$" to be "$+$", this equation reduces to $(2)$, so we have successfully generalized OP's parabolas.
However, the vertices of the generalized conic have coordinates $(\overline{x}, \overline{y}) := (h, k\pm_1 a)$, where "$\pm_1$" is independent of "$\pm$". That is,
$$\overline{x} = \pm \frac{p \sin\theta \cos^2\theta}{
 \sqrt{1 - e^2 \sin^2\theta}} \qquad \overline{y} = \pm \frac{p \cos^2\theta \left(-\cos\theta \pm_1 \sqrt{1 - e^2 \sin^2\theta}\right)}{(1 - e^2)\sqrt{1 - e^2 \sin^2\theta}} \tag{9}$$
Eliminating the parameter $\theta$ yields

$$2 p y \left(x^2 - y^2(1-e^2)\right)^2 \pm\left(x^2 + y^2 (1-e^2)\right) \left(x^2 + y^2 (1-e)^2\right) \left(x^2 + y^2 (1+e)^2\right) = 0 \tag{10}$$

which, in general, does not itself describe a conic. (For $e=0$, this reduces to ellipse equation $(4)$, with an extra factor of $x^4$.)

Note that, since the parabola's eccentricity is $1$, it could be hiding just-about anywhere in a formula, waiting to be generalized.
As a specific example, motivated by a quantity appearing often in the above analysis, we could consider the "third" condition to be that the semi-latus rectum is not $p \cos^2\theta$, but rather $p (1 - e^2\sin^2\theta)$. The resulting vertex locus here is also generally not a conic:

$$\left(\;x^2 - (\pm 1)(\pm_1 1)\,2 p y + 4 y^2\;\right) \left(\;x^2 + y^2(1- e^2)\right) = y^2 (1 - e^2) \left(3 x^2 + 3 y^2 + e^2 y^2\right) \tag{11}$$


It's possible ---even likely--- that some other generalization gives a more-satisfying result. Indeed, one could work backwards, replacing the latus-rectum condition with an assumption that the vertex locus is some conic (or whatever curve might be desired). Such exploration is left to the reader.
A: Here's a better generalization than in my previous answer, using a different interpretation of that answer's equation $(3)$:

$$\left(\;y - \tfrac12\sin^2\theta\;\right) = -\frac{1}{2p\cos^2\theta}\left(\;x-\tfrac12\sin 2\theta\;\right)^2 \tag{1}$$

This time around, instead of interpreting $p\cos^2\theta$ as the semi-latus-rectum, we recognize it as twice the vertex-to-focus distance. This places the focus at
$$(\hat{x},\hat{y}) = \left(\;\tfrac12 p\sin 2\theta, \tfrac12 p\sin^2\theta - \tfrac12p\cos^2\theta\;\right) = \left(\;\tfrac12 p\sin 2\theta,-\tfrac12p\cos 2\theta\;\right) \tag{2}$$
which is on a circle of radius $r := \tfrac12p$. Let's use that as an alternative "third condition" from before; since the distance from center to focus is $ae$, we can replace the earlier equation $(6a)$ with
$$h^2 + ( k + ae )^2 = r^2 \tag{3}$$
Solving the new system for $a$, $h$, $k$ gives (upon discarding a couple of extraneous cases)
$$
a = \frac{\sigma r m (m + e \cos\theta)}{1 - e^2} \qquad 
h = \sigma r \sin\theta (m + e \cos\theta) \qquad
k = - \frac{\sigma r \cos\theta (m + e \cos\theta)}{1 - e^2} \tag{4}$$
where $m := \sqrt{1-e^2\sin^2\theta}$ and $\sigma:=\pm 1$, so that the generalized conic has this equation:
$$x^2 + y^2 (1- e^2) + 2 \sigma r ( x \sin\theta - y \cos\theta ) \left( m + e \cos\theta \right) = 0 \tag{5}$$
The foci have coordinates
$$(\hat{x}, \hat{y}) = (h, k + \sigma_1 a) = \left(\; \sigma r\sin\theta(m+e\cos\theta), \;\frac{\sigma r (m + e \cos\theta) (\cos\theta -\sigma_1 m)}{1 - e^2}\;\right) \tag{6}$$
that satisfy 
$$x^2 + y^2( 1 + \sigma_1 e)^2 + 2 \sigma \sigma_1 y r ( 1 + \sigma_1 e )  = 0 \tag{7}$$
That is,

$$\frac{x^2}{r^2} + \frac{(1 + \sigma_1 e)^2}{r^2} \left(y + \frac{\sigma\sigma_1 r}{1 + \sigma_1 e}\right)^2 = 1 \tag{8}$$

which indicates that each vertex of our conic has an elliptical locus. For $\sigma=1$, the horizontal axis is major; for $\sigma = -1$, that axis is minor. The corresponding eccentricities are

$$\sigma_1 = 1 \;:\;\frac{\sqrt{e(2+e)}}{1+e} \qquad \sigma_1 = -1\;:\; \sqrt{e(2-e)} \tag{9}$$

that, as with OP's original scenario, are independent of $r$ (or $p$, or $v$ and $g$). For $e=1$, these are $\sqrt{3}/4$ and $1$; the first confirms OP's observation, while the second is extraneous here.) $\square$
Here are some animated illustrations, showing scenarios for families of conics with eccentricity $0$, $0.5$, $0.999$, and $1.5$. (In the last case, observe that no drawing occurs while $1-e^2\sin^2\theta$ is negative.)




A: Here's yet-another pass at the problem, drawing an even-more-general conclusion: 

$$\text{There are many ellipses!}$$

I'm posting separately because I'll be using a different parameterization to help streamline things.

The jumping-off point is the observation made in this previous answer that the focus-locus for the variable parabola in OP's scenario is a circle. So, for a vertically-oriented conic, define a generic "upper" focus, $F$ as
$$F = r (-\sin\theta, \cos\theta) \tag{1}$$
(This $\theta$ is different from the one in my other answers. Also, $\theta=0$ corresponds to the top of the circle.) For an ellipse/hyperbola with eccentricity $e$ and major/transverse radius $a$, the focus-to-focus distance is $2ae$, so the other focus is $G=F - 2a e(0,1)$. For the ellipse (or the lower branch of the hyperbola) to contain the origin, we have
$$|OF| \pm |OG| = 2a \quad\to\quad 4 a^2 e^2 + r^2 - 4 a e r \cos\theta = ( 2 a - r )^2 \quad\to\quad a = r\,\frac{1 - e \cos\theta}{1 - e^2} \tag{2}$$
The resulting conic, then, has this equation:
$$x^2 + y^2\left(1-e^2\right) + 2 x r \sin\theta + 2 y r ( e - \cos\theta )  = 0 \tag{3}$$
(where $e=1$ is allowed). Noting that the semi-latus-recum is $1-e\cos\theta$, we can parameterize this conic with the help of the focus-at-origin polar form to get
$$( x, y ) =  F + r\;\frac{1-e\cos\theta}{1+e\cos\phi}\;\left( -\sin\phi, \cos\phi \right) \tag{4}$$
For a given $\phi$ the parameterization matches "similar" points across the $\theta$-family of conics: $\phi=0$ gives all "upper" vertices; $\phi=\pi$, all "lower" vertices; $\phi=\pm\pi/2$, all endpoints of the latus rectum through $F$; etc. Eliminating $\theta$, we find that the locus of each collection "similar" points is a conic:

$$x^2 + y^2 \left( 1 + e^2 - 2 e \cos\phi \right) - 2 x y e \sin\phi + 2 x r \sin\phi - 2 r y \left( e + \cos\phi \right)  = 0 \tag{5}$$

Since
$$ \left(- 2 e \sin \phi \right)^2 - 4 (1) \left(1+e^2-2 e\cos\phi\right) = -4 (1 + e \cos\phi )^2 < 0 \tag{6}$$
each conic is specifically an ellipse. The eccentricity, $\hat{e}$, is given by
$$\hat{e}^2 = \frac{4 e \sqrt{4 + 4 e\cos\phi + e^2}}{\left(e + \sqrt{4 + 4e \cos\phi + e^2} \right)^2} \tag{7}$$
In particular, we have
$$\begin{align}
\phi=0 &\;\to\; \hat{e}^2 = e\,\frac{2 + e}{(1 + e)^2}\\
\phi = \pi &\;\to\; \hat{e}^2 = 4 e \frac{|2 - e|}{\left(e + |2 - e|\right)^2} = \begin{cases}
e (2 - e) & e \leq 2 \\[4pt]
e \dfrac{e - 2}{(e-1)^2} & e > 2\end{cases} \\
\phi = \pm \tfrac12\pi &\;\to\; \hat{e}^2 = 4 e \frac{\sqrt{4 + e^2}}{\left(e + \sqrt{4 + e^2}\right)^2}
\end{align} \tag{8}$$
Here are a few animations, for $e=0$, $e=0.5$, $e=1$, $e=1.5$, showing the ellipses for the four multiple-of-a-right-angle values of $\phi$.




