Question: Projectiles are fired with initial speed $v$ and variable launch angle $0< \alpha < \pi$.
Choose a coordinate system with the firing position at the origin. For each value of $\alpha$ the trajectory will follow a parabolic arc with apex at $(x,y)$ where both $x$ and $y$ depend on $\alpha$. Show that:
$$ x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2 = \frac{v^2}{4g^2} $$
and hence the points of maximum heigh of the trajectories lie on an ellipse.
Requires some basic understanding of physics.
What I have attempted:
As the initial velocity is $v$ , resolving the components
$$ v_x = v\cos(\alpha) $$
$$ v_y = v\sin(\alpha) $$
To find the time to apex (vertex) at $(x,y)$ would be using the formula
$$ v_f = v_i + at $$
$v_f$ at the top is 0 and $a=-g$ and $v_i = v\sin(\alpha)$ hence
$$ t_{top} = \frac{ v\sin(\alpha)}{g} $$
and by using the formula $$ y = v_yt + \frac{1}{2}a_yt^2 $$
$$ y = \frac{1}{2}g t^2 $$
$$ y_{apex} = \frac{v^2\sin^2(\alpha)}{2g} $$
And as horizontal movement is at constant velocity
$$ x = v_x \cdot t $$
$$ x= v\cos(\alpha) \cdot \frac{ v\sin(\alpha)}{g} $$
$$ x = \frac{v^2 \sin(\alpha)\cos(\alpha)}{g} $$
$$ x_{apex} = \frac{v^2 \sin(2\alpha)}{2g} $$
Now I am stuck how do I prove the equation....
