[Q1] Find the equation of the curve formed by the highest points of all possible parabolic trajectories.

The position of a projectile launched from the origin with initial speed $v_0$ at an angle $\theta$ is given by:

$$x(t) = (v_0 \cos\theta) t$$ $$y(t) = (v_0 \sin\theta) t - \frac{1}{2}gt^2$$

The corresponding velocity components are:

$$v_x(t) = v_0 \cos\theta$$ $$v_y(t) = v_0 \sin\theta - gt$$

The apex of the trajectory is the highest point, where the vertical component of velocity is zero. Let $t_a$ be the time to reach the apex.

$$v_y(t_a) = v_0 \sin\theta - gt_a = 0 \implies t_a = \frac{v_0 \sin\theta}{g}$$

The coordinates of the apex $(x_a, y_a)$ are found by substituting $t_a$ into the position equations.

$$x_a = (v_0 \cos\theta) \left( \frac{v_0 \sin\theta}{g} \right) = \frac{v_0^2 \sin\theta \cos\theta}{g}$$ $$y_a = (v_0 \sin\theta) \left( \frac{v_0 \sin\theta}{g} \right) - \frac{1}{2}g \left( \frac{v_0 \sin\theta}{g} \right)^2 = \frac{v_0^2 \sin^2\theta}{2g}$$

To find the locus of these apexes, we eliminate the parameter $\theta$. Let $(x, y)$ be a point on the locus, so $x=x_a$ and $y=y_a$. From the equation for $y_a$:

$$\sin^2\theta = \frac{2gy}{v_0^2}$$

Using the identity $\cos^2\theta = 1 - \sin^2\theta$:

$$\cos^2\theta = 1 - \frac{2gy}{v_0^2}$$

Now, square the equation for $x_a$:

$$x^2 = \frac{v_0^4}{g^2} \sin^2\theta \cos^2\theta$$

Substitute the expressions for $\sin^2\theta$ and $\cos^2\theta$ in terms of $y$:

$$x^2 = \frac{v_0^4}{g^2} \left( \frac{2gy}{v_0^2} \right) \left( 1 - \frac{2gy}{v_0^2} \right)$$

Simplifying the expression gives the equation of the locus:

$$x^2 = \frac{2v_0^2 y}{g} \left( 1 - \frac{2gy}{v_0^2} \right)$$ $$x^2 = \frac{2v_0^2}{g}y - 4y^2$$

Rearranging this yields the final equation:

$$x^2 + 4y^2 - \frac{2v_0^2}{g}y = 0$$

This is the equation of an ellipse centered at $(0, v_0^2/4g)$.