Editing Projectile motion (section)

== Kinematic quantities ==
In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. This is the principle of ''compound motion'' established by [[Galileo]] in 1638,<ref>Galileo Galilei, ''[[Two New Sciences]]'', Leiden, 1638, p.249</ref> and used by him to prove the parabolic form of projectile motion.<ref>Nolte, David D., Galileo Unbound (Oxford University Press, 2018) pp. 39-63.</ref>
[[File:Compound Motion.gif|thumb|The horizontal and vertical components of a projectile's velocity are independent of each other.]]
A ballistic trajectory is a parabola with homogeneous acceleration, such as in a space ship with constant acceleration in absence of other forces. On Earth the acceleration changes magnitude with altitude as <math display="inline">g(y)=g_0/(1+y/R)^2</math> and direction (faraway targets) with latitude/longitude along the trajectory. This causes an [[Ellipse|elliptic]] trajectory, which is very close to a parabola on a small scale. However, if an object was thrown and the Earth was suddenly replaced with a [[black hole]] of equal mass, it would become obvious that the ballistic trajectory is part of an elliptic [[orbit]] around that "black hole", and not a parabola that extends to infinity. At higher speeds the trajectory can also be circular ([[Astronautics|cosmonautics]] at [[Low Earth orbit|LEO]]?, [[Geostationary orbit|geostationary]] satellites at 5<math display="inline">\frac{5}{6}</math> R), parabolic or [[Hyperbola|hyperbolic]] (unless distorted by other objects like the Moon or the Sun).

In this article a ''homogeneous'' gravitational acceleration <math display="inline">(g=g_0)</math> is assumed.

=== Acceleration ===
Since there is acceleration only in the vertical direction, the velocity in the horizontal direction is constant, being equal to <math> \mathbf{v}_0 \cos\theta </math>. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to <var>g</var>.{{NoteTag|<var>g</var> is the [[Standard gravity|acceleration due to gravity]]. (<math>9.81\,\mathrm{m/s^2}</math> near the surface of the Earth).}} The components of the acceleration are: 
: <math> a_x = 0 </math>,
: <math> a_y = -g </math>.*

''*The y acceleration can also be referred to as the force of the earth'' <math display="inline">(-F_g/m)</math> ''on the object(s) of interest.''

=== Velocity ===
Let the projectile be launched with an initial [[velocity]] <math>\mathbf{v}(0) \equiv \mathbf{v}_0 </math>, which can be expressed as the sum of horizontal and vertical components as follows:
:<math> \mathbf{v}_0 = v_{0x}\mathbf{\hat x} + v_{0y}\mathbf{\hat y} </math>.
The components <math> v_{0x} </math> and <math> v_{0y} </math> can be found if the initial launch angle θ is known:
:<math> v_{0x} = v_0\cos(\theta)</math>,
:<math> v_{0y} = v_0\sin(\theta)</math>

The horizontal component of the [[velocity]] of the object remains unchanged throughout the motion. The vertical component of the velocity changes linearly,{{NoteTag|decreasing when the object goes upward, and increasing when it goes downward}} because the acceleration due to gravity is constant. The accelerations in the <var>x</var> and <var>y</var> directions can be integrated to solve for the components of velocity at any time <var>t</var>, as follows:
: <math> v_x = v_0 \cos(\theta) </math>,
: <math> v_y = v_0 \sin(\theta) - gt </math>.

The magnitude of the velocity (under the [[Pythagorean theorem]], also known as the triangle law):
: <math> v = \sqrt{v_x^2 + v_y^2 } </math>.

=== 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>.

=== Displacement in polar coordinates ===
The parabolic trajectory of a projectile can also be expressed in polar coordinates instead of [[Cartesian coordinate system|Cartesian]] coordinates. In this case, the position has the general formula
:<math> r( \phi ) = \frac{2v_0^2 \cos^2\theta}{|g|} \left(\tan\theta\sec\phi -\tan\phi\sec\phi \right)  </math>.
In this equation, the origin is the midpoint of the horizontal range of the projectile, and if the ground is flat, the parabolic arc is plotted in the range <math> 0 \leq \phi \leq \pi </math>. This expression can be obtained by transforming the Cartesian equation as stated above by <math> y = r \sin\phi </math> and <math> x = r \cos\phi </math>.