Show that the trajectory of a particle that is launched at an angle is parabolic?

Jul 11, 2017

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

$\left.\begin{matrix}v = u + a t & \text{ where " & s="displacement "(m) \\ s=ut+1/2at^2 & \null & u="initial speed "(ms^-1) \\ s=1/2(u+v)t & \null & v="final speed "(ms^-1) \\ v^2=u^2+2as & \null & a="acceleration "(ms^-2) \\ s=vt-1/2at^2 & \null & t="time } \left(s\right)\end{matrix}\right.$

Consider a projectile launched at an angle $\theta$ with initial speed $u$. We ignore air resistance, and model the projectile as a particle with all mass concentrated at a point.

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with a=0).

The projectile will travel a distance $x$ in time $t$, we must resolve the initial speed in the horizontal direction

$\left\{\begin{matrix}s = & x & m \\ u = & u \cos \theta & m {s}^{-} 1 \\ v = & \text{Not Required} & m {s}^{-} 1 \\ a = & 0 & m {s}^{-} 2 \\ t = & t & s\end{matrix}\right.$

So applying $s = u t + \frac{1}{2} a {t}^{2}$ we get

$x = u \setminus \cos \theta \setminus t$

Vertical Motion

The projectile travels under constant acceleration due to gravity. Considering upwards as positive, and again resolving the initial speed (vertically this time):

$\left\{\begin{matrix}s = & y & m \\ u = & u \sin \theta & m {s}^{-} 1 \\ v = & \text{Not Required} & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & t & s\end{matrix}\right.$

Applying $s = u t + \frac{1}{2} a {t}^{2}$ we have:

$y = u \setminus \sin \theta \setminus t + \frac{1}{2} \left(- g\right) {t}^{2}$
$\setminus \setminus = u \setminus \sin \theta \setminus t - \frac{1}{2} g {t}^{2}$

Substituting for $t$ from the first equation into the second we get:

$y = u \setminus \sin \theta \setminus \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g {\left(\frac{x}{u \cos \theta}\right)}^{2}$
$\setminus \setminus = \left(\tan \theta\right) x - \frac{g}{2 {u}^{2} {\cos}^{2} \theta} \setminus {x}^{2}$
$\setminus \setminus = a x - b {x}^{2} \setminus \setminus$, say, where $a , b$ are constants.

Which is the trajectory of a parabola.