# 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 $V_{0}$ 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:

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

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

We obtain $t$ from  and we replace it in  to find $y(x)$ :

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

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

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

3. MAXIMUM HEIGHT

To find the maximum height $y_{max}$ we replace $t_{imp}$ in the :

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