# Projectile motion: Proving:$x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2 = \frac{v^2}{4g^2}$

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....

We have: $$4\left(y-\frac{v^2}{4g}\right)^2=4\left(\frac{v^2\sin^2(\alpha)}{2g}-\frac{v^2}{4g}\right)^2=\frac{v^4\sin^4(\alpha)}{g^2}-\frac{v^4\sin^2(\alpha)}{g^2}+\frac{v^4}{4g^2}$$ $$=\left(\frac{v^4\sin^2\alpha}{g^2}\right)(\sin^2\alpha-1)+\frac{v^4}{4g^2}$$ $$=-\left(\frac{v^4\sin^2\alpha}{g^2}\right)(1-\sin^2\alpha)+\frac{v^4}{4g^2}$$ $$=-\left(\frac{v^4\sin^2\alpha}{g^2}\right)\cos^2\alpha+\frac{v^4}{4g^2}$$ and $$x^2= \frac{v^4 \sin^2\alpha\cos^2\alpha}{g^2}$$ so $$x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2=\frac{v^4 \sin^2\alpha\cos^2\alpha}{g^2}-\frac{v^4\sin^2\alpha\cos^2\alpha}{g^2}+\frac{v^4}{4g^2}$$ $$=\frac{v^4}{4g^2}$$