Editing Projectile motion (section)

=== Displacement ===
[[File:Ferde hajitas3.svg|thumb|250px|Displacement and coordinates of parabolic throwing]]
At any time <math> t </math>, the projectile's horizontal and vertical [[displacement (vector)|displacement]] are:
:<math> x = v_0 t \cos(\theta) </math>,
:<math> y = v_0 t \sin(\theta) - \frac{1}{2}gt^2 </math>.
The magnitude of the displacement is:
:<math> \Delta r=\sqrt{x^2 + y^2 } </math>.

Consider the equations,
:<math> x = v_0 t \cos(\theta) </math> and <math>y = v_0 t\sin(\theta) - \frac{1}{2}gt^2</math>.<ref>{{Cite book |last=Stewart |first=James |title=Calculus: Early Transcendentals |last2=Clegg |first2=Dan |last3=Watson |first3=Saleem |date=2021 |publisher=Cengage |isbn=978-1-337-61392-7 |edition=Ninth |location=Boston, MA |pages=919}}</ref>
If <var>t</var> is eliminated between these two equations the following equation is obtained:
:<math> y = \tan(\theta) \cdot x-\frac{g}{2v^2_{0}\cos^2 \theta} \cdot x^2=\tan\theta \cdot x \left(1-\frac{x}{R}\right). </math>
Here R is the [[range of a projectile]].

Since <var>g</var>, <var>θ</var>, and <var>v<sub>0</sub></var> are constants, the above equation is of the form 
:<math> y=ax+bx^2 </math>,
in which <var>a</var> and <var>b</var> are constants. This is the equation of a parabola, so the path is parabolic. The axis of the parabola is vertical.

If the projectile's position (x,y) and launch angle (&theta; or &alpha;) are known, the initial velocity can be found solving for<var> v<sub>0</sub></var> in the afore-mentioned parabolic equation:

:<math> v_0 = \sqrt{{x^2 g} \over {x \sin 2\theta - 2y \cos^2\theta}} </math>.