I am trying to create a function that will generate a parabola that describes projectile motion.
Here are my inputs:
I need an output for an equation of a parabola that fits this motion.
On
EDIT1:
Write down vertical acceleration equilibrium and integrate two times,remembering to include vertical component of projectile velocity $V$ shot at $ \alpha $ to x-axis as follows.
The accelerations in x- and y-(vertical) directions are:
\begin{align} \ddot x(t) &= 0\\ \ddot y(t) &= -g \end{align} First integration for velocity with components of initial $V$ components as : \begin{align} \dot x(t) &= V \cos \alpha \\ \dot y(t) &= V \sin \alpha -g \, t \\ \end{align} Second integration is for position starting at origin with boundary condition ( x=0, y=0 ): \begin{align} \dot x(t) &= V \cos \alpha \,t \\ \dot y(t) &= V \sin \alpha \,t -g t^2/2 \, \\ \end{align}
Now eliminate $t$ to find trajectory as a parabola reaching a maximum point before hitting ground.
$$ y = x \tan \alpha - g x^2 / 2 /(V \cos \alpha)^2 $$
It is possible to find maximum range and height using using above relations. Maximum range occurs at $ \alpha = 45^0 $ and of course maximum height $ V^2 / 2g $ when shot vertically up.
It is an interesting fact I found that as $\alpha $ is varied, the envelope of all parabolas is another parabola.
By Newton's second law $F = ma = m(0,-g)$, which es equivalent to the set of equations: \begin{align} \ddot x(t) &= 0\\ \ddot y(t) &= -g. \end{align} After integrating these, we get that: \begin{align} x(t) &= x_0+v_{x}t\\ y(t) &= y_0+v_yt-\frac{gt^2}{2}, \end{align} where $(x_0,y_0)$ is the initial position of the object and $(v_x,v_y)$ its initial velocity. These equations define a parabola.
If you want to get an equation of the form $y = f(x)$, use that $t= \frac{x-x_0}{v_x}$, then: $$ y = y_0+v_y\frac{x-x_0}{v_x} - \frac{g(x-x_0)^2}{2v_x^2}, $$ where $x\in[x_0,x_1]$ and $x_1$ is where the projectile reaches the ground again.