# Parabolic motion

The parabolic motion is the union of two simultaneous and independent straight motions, one uniform and one constantly accelerated.

Projectile motion with initial velocity ${\displaystyle V_{0}}$ and an angle θ In this case the projectile is constantly accelerated by the gravitational acceleration along the y axis, on the other axis the motion is uniform as there are no forces or accelerations.

INITIAL CONDITIONS:

{\displaystyle \left\{{\begin{aligned}&X_{0}=0\\&y_{0}=0\\\end{aligned}}\right.}
{\displaystyle \left\{{\begin{aligned}&V_{0x}=V_{0}\cos \theta \\&V_{0y}=V_{0}\sin \theta \\\end{aligned}}\right.}

SCOMPOSITION OF MOTION:

{\displaystyle \left\{{\begin{aligned}&x(t)=x_{0}+V_{0x}t\\&y(t)=V_{0y}t-{\frac {1}{2}}gt^{2}\\\end{aligned}}\right.}
{\displaystyle \left\{{\begin{aligned}&x(t)=V_{0}\cos \theta \cdot t\qquad [1]\\&y(t)=V_{0}sin\theta \cdot t-{\frac {1}{2}}gt^{2}\qquad [2]\\\end{aligned}}\right.}

1. TAJECTORY

We obtain ${\displaystyle t}$ from [1] and we replace it in [2] to find ${\displaystyle y(x)}$:

${\displaystyle t={\frac {x}{V_{0}\cos \theta }};}$
${\displaystyle y(x)=V_{0}\sin \theta {\frac {x}{V_{0}\cos \theta }}-{\frac {g}{2}}\cdot {\frac {x^{2}}{V_{0}^{2}\cos ^{2}\theta }};}$
${\displaystyle y(x)=\tan \theta x-{\frac {g}{2}}{\frac {x^{2}}{V_{0}^{2}\cos ^{2}\theta }}}$

2. RANGE

We impose ${\displaystyle y(x)=0}$ in order to find the total horizontal displacement:

${\displaystyle \tan \theta x-{\frac {g}{2}}{\frac {x^{2}}{V_{0}^{2}\cos ^{2}\theta }}=0;}$
${\displaystyle {\frac {V_{0}\sin \theta x}{V_{0}\cos \theta }}-{\frac {gx^{2}}{2V_{0}^{2}\cos ^{2}\theta }}=0}$
${\displaystyle {\frac {2\sin \theta \cos \theta V_{0}^{2}x-gx^{2}}{2V_{0}^{2}\cos ^{2}\theta }}}$
As ${\displaystyle 2\sin \theta \cos \theta =\sin 2\theta :}$
${\displaystyle x(V_{0}^{2}\sin 2\theta -gx)=0;}$
we find
${\displaystyle x=0}$
which is the position of launch and
${\displaystyle x={\frac {\sin 2\theta \cdot V_{0}^{2}}{g}}}$
which is the range.

3. MAXIMUM HEIGHT

To find the maximum height ${\displaystyle y_{max}}$ we replace ${\displaystyle t_{imp}}$ in the [2]:

${\displaystyle y_{max}={\frac {V_{0y}^{2}}{2g}}={\frac {V_{0}^{2}\sin ^{2}\theta }{2g}}}$