P02-030 – Equation of the Path y = f(x)
Equation of the path $y\left(x\right)$
The kinematic equations $x\left(t\right)$ and $y\left(t\right)$ below give the $x$ and $y$ position of the projectile as a function of time.
\begin{aligned}
x\left(t\right)&=v_0{\mathrm{cos} \left(\theta \right)\ }t+x_0 \\
y\left(t\right)&=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t+y_0
\end{aligned}They can also be rearranged to directly give the height $y$ of the projectile in terms of its position $x$ which yields the equation of the parabolic path the projectile follows.
This process, also known as ”eliminating the parameter” ($t$ being the parameter), can be systematically done by expressing $t$ in terms of $x$ as done below
x=v_0{\mathrm{cos} \left(\theta \right)\ }t+x_0\ \ \ \ \Rightarrow \ \ \ \ \ t=\frac{x-x_0}{v_0{\mathrm{cos} \left(\theta \right)\ }}The expression derived for $t$ can then be substituted into $y\left(t\right)$ to yield $y\left(x\right)$
\begin{aligned}
y\left(x\right)&=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t+y_0 \\
\\
&=-\frac{1}{2}g{\left(\frac{x-x_0}{v_0{\mathrm{cos} \left(\theta \right)\ }}\right)}^2+v_0{\mathrm{sin} \left(\theta \right)\ }\left(\frac{x-x_0}{v_0{\mathrm{cos} \left(\theta \right)\ }}\right)+y_0 \\
\\
&=-\frac{g{\left(x-x_0\right)}^2}{{2v}_0{{\mathrm{cos}}^2 \left(\theta \right)\ }}+{\mathrm{tan} \left(\theta \right)\ }\left(x-x_0\right)+y_0
\end{aligned}The equation of the parabolic path followed by the projectile is therefore given by
\boxed{y\left(x\right)=-\frac{g{\left(x-x_0\right)}^2}{{2v}_0{{\mathrm{cos}}^2 \left(\theta \right)\ }}+{\mathrm{tan} \left(\theta \right)\ }\left(x-x_0\right)+y_0}Note: special case if $x_0=y_0=0$
In this common case, the expression $y\left(x\right)$ simplifies to
\boxed{y\left(x\right)=-\frac{gx^2}{{2v}_0{{\mathrm{cos}}^2 \left(\theta \right)\ }}+{\mathrm{tan} \left(\theta \right)\ }x+y_0}