Highest Points

At any time $t$ , the projectile's horizontal and vertical displacement are

$\begin{aligned} x&=vt\cos\theta,\\ y&=vt\sin\theta-\frac 12gt^2. \end{aligned}$

The increase in height will last until $v_y=0$ , that is, $0=v\sin\theta-g t$ . Thus, time to reach the maximum height is given by

$t=\frac {v\sin\theta}g.$

Substituting this value into the equations above, we get coordinates of highest point

$\begin{aligned} X&=\frac {v^2\sin2\theta}{2g},\\ Y&=\frac {v^2\sin^2\theta}{2g}. \end{aligned}$

If $\theta$ is eliminated between these two equtions, the following eqution is obtained

$\frac {X^2}{\left(\frac {v^2}{2g}\right)^2}+\frac {\left(Y-\frac {v^2}{4g}\right)^2}{\left(\frac {v^2}{4g}\right)^2}=1.$

Therefore, the set of all highest points is a ellipse centered at $(0,\frac {v^2}{4g})$ with semi-major and semi-minor axes $\frac {v^2}{2g}$ and $\frac {v^2}{4g}$ respectively. Its area is

$A=\pi\left(\frac {v^2}{2g}\right)\left(\frac {v^2}{4g}\right)=\left(\frac \pi8\right)\left(\frac {v^4}{g^2}\right),$

making the answer $\boxed{\frac\pi8}.$

I got $X$ and $Y$ as Brian did (those formulas can be found in introductory physics texts as well), and then let $A=\int_{\theta =0}^{\pi} X dY =\frac{v^4}{4g^2} \int_{\theta =0}^{\pi} \sin^2(2\theta) d\theta= \frac{v^4\pi}{8g^2}$ The answer is $\alpha=\frac{\pi}{8}\approx \boxed {0.3927}$