Answer
$y = \sqrt{3}~x-\frac{x^2}{256}+8$
The rocket follows a path that is a parabola.
Work Step by Step
$v = 128~ft/s$
$\theta = 60^{\circ}$
$h = 8$
We can find an expression for $x$ in terms of $t$:
$x = (v~cos~\theta)(t)$
$x = (128~cos~60^{\circ})(t)$
$x = 64~t$
Then: $t = \frac{x}{64}$
We can find an expression for $y$ in terms of $t$:
$y = (v~sin~\theta)(t)-16~t^2+h$
$y = (128~sin~60^{\circ})(t)-16~t^2+8$
$y = 64~\sqrt{3}~t-16~t^2+8$
We can replace $t$ with the value $t = \frac{x}{64}$:
$y = 64~\sqrt{3}~t-16~t^2+8$
$y = 64~\sqrt{3}~(\frac{x}{64})-16~(\frac{x}{64})^2+8$
$y = \sqrt{3}~x-\frac{x^2}{256}+8$
This equation is an equation for a parabola, so the rocket follows a path that is a parabola.
