Parabolic trajectory

A trajectory is known as parabolic if the movement of a body in space describes a Parabole.

The parabolic movement is carried out when a Projectile is subjected at an initial speed and only the Accélération of the Pesanteur. An example running of parabolic movement is the drawn Obus since a gun.

The discovery of the parabolic trajectory is allotted to Galileo in 1638. Certain historians of sciences think that it was largely influenced by the artists of its time who could represent the trajectory of the water of the fountains.

Examples

When one launches an object in the air, except the case where it was launched rigorously to the vertical to the top, its trajectory is a curve which one can compare to a parabola. For example, the shooting of a ball of gun or a boules ball describes a quasi-parabolic trajectory. The Comet S pass in the vicinity of Sun or Earth on an orbit " parabolic ". If a plane carries out a parabolic trajectory, then the passengers embarked are in Impesanteur.

Study of the trajectory of a projectile

The movement of an object subjected to a field of uniform gravity (in the absence of frictions) is a parabolic trajectory (Balistique).

That is to say a specific body supposed of Mass m , studied in a reference mark (O, X, there, Z) , supposed galiléen Z being the vertical, directed upwards. This body is placed in a field of gravity, the acceleration of gravity is G . The body is launched since the point ( x₀, y₀, z₀ ) with an initial speed: $\ vec V_0= \ begin {pmatrix} V_x \ \ 0 \ \ V_z \ end {pmatrix}$

It is supposed here that there is no component speed along the axis $\ vec y$ , all the movement thus takes place in a Plan parallel with the plan ( xOz ) . One notes T time.

Resolution of the equation

The only force to which the body is subjected is the Gravité (one can refine the problem by adding for example friction due to the air). The only acceleration printed with the body is thus the acceleration of gravity.

$\ vec has = \ begin {pmatrix} 0 \ \ 0 \ \ - G \ end {pmatrix}$

To deduce speed from it, it is enough to integrate acceleration.

$\ vec V = \ begin {pmatrix} C1 \ \ C2 \ \ - G \, t+C3 \ end {pmatrix}$

C1, C2 and C3 is constant of integration, given by the initial conditions. Indeed with T = 0 , $\ vec V = \ vec V_0$ , is $\ begin {pmatrix} C1 \ \ C2 \ \ - G \ times 0+C3 \ end {pmatrix} = \ begin {pmatrix} V_x \ \ 0 \ \ V_z \ end {pmatrix}$ ,

from where C1 = V_x, C2 = 0 and C3=V_z .

One thus has

$\ vec V (T) = \ begin {pmatrix} V_x \ \ 0 \ \ - G \, t+V_z \ end {pmatrix}$

To obtain the equation of the Trajectory, speed should be integrated.

$\ overrightarrow {OM} (T) = \ begin {pmatrix} V_x \, t+C4 \ \ C5 \ \ - {1 \ over 2} \, G \, t^2+V_z \, t+C6 \ end {pmatrix}$

C4, C5 and C6 is (again) constants of integration which will be given using the initial conditions.

With T = 0 , $\ overrightarrow {OM} (0) = \ overrightarrow {OM_0}$

Thus $\ begin {pmatrix} V_x \ times 0+C4 \ \ C5 \ \ - {1 \ over 2} \, G \ times 0^2+V_z \ times 0+C6 \ end {pmatrix} = \ begin {pmatrix} x_0 \ \ y_0 \ \ z_0 \ end {pmatrix}.$
from where $\ overrightarrow {OM} (T) = \ begin {pmatrix} V_x \, t+x_0 \ \ y_0 \ \ - {1 \ over 2} \, G \, t^2+V_z \, t+z_0 \ end {pmatrix}$

Equation of the trajectory

One can give the equation in the form Z = F (X) ( Z is a function of X ) by replacing T in the equation of Z by the expression that one draws some in the equation from X , that is to say $t= {x-x_0 \ over V_x}$

One thus obtains: $z (X) = - {1 \ over 2} \, G \, \ left ({x-x_0 \ over V_x} \ right) ^2+V_z \, {x-x_0 \ over V_x} +z_0$

$Z (X) = - {1 \ over 2} \, {G \ over V_x^2} \, x^2+ \ left ({V_z \ over V_x} - {G \ over V_x^2} x_0 \ right) \, X {1 \ over 2} \, {G \ over V_x^2} \, x_0^2- {V_z \ over V_x} \, x_0+z_0$

The equation of this movement clearly shows the Parabole which gives its name to this movement. This equation also makes it possible to withdraw several information useful such as for example the places or the projectile touches the ground (to solve the equation Z (X) = 0 ).

Graph

Here $\ phi$ is the angle which the initial Flight Path Vector with the horizontale&thinsp forms;:
$\ phi = arctan (\ frac {V_y} {V_x})$

Notes and references of the article

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