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....


1 Answer 1


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 $$x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2=\frac{v^4 \sin^2\alpha\cos^2\alpha}{g^2}-\frac{v^4\sin^2\alpha\cos^2\alpha}{g^2}+\frac{v^4}{4g^2}$$ $$=\frac{v^4}{4g^2}$$


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