trouver un angle avec l'equation de trajectoire d'un projectile

Bonjour, j'ai un probleme avec ma conception d'un projet de cours j'aimerais calculé l'angle adequat pour mon projectile avec un x0 et y0 en fonction de x mais matlab me donne aucune reponse. voici ce que je recois comme resultat. merci

[Torsten]

You can compute theta as a function of x and t, but how should it be possible to compute theta solely as a function of x ?

Vous pouvez calculer thêta en fonction de x et t, mais comment serait-il possible de calculer thêta uniquement en fonction de x ?

[Franck Aurel]

J ai deja compris t dans l equation en trouvant t en fonction de y et theta et vu que a l arrivée y=0 on a juste y0 dans l equation donc cette forme d equation prends en compte t x et y deja

I already included t in the equation by finding t as a function of y and theta and seeing that at the end y=0 we just have y0 in the equation so this form of equation takes into account t x and y already

Answer 1

[Torsten]

x0 = 0.08;
y0 = 0.05;
v0 = 2;
g = 9.81;
syms t x theta
eqn1 = x == v0*t*cos(theta) + x0;
eqn2 = 0 == v0*t*sin(theta) - 0.5*g*t^2 + y0;
sol = solve([eqn1,eqn2],[t,theta],'MaxDegree',4);
simplify(sol.t)

ans = 

●

simplify(sol.theta)

ans = 

[Franck Aurel]

J’essaie cela et je vous reviens

I'll try this and get back to you.

[Torsten]

Of course, x must be chosen such that the point (x,0) can be reached by a certain angle for theta.

If this is the case, I suspect that two solutions will give imaginary values and the third solution will give a negative time so that only one physically senseful solution will remain.

Bien sûr, x doit être choisi de telle sorte que le point (x, 0) puisse être atteint sous un certain angle pour thêta.

Si tel est le cas, je suppose que deux solutions donneront des valeurs imaginaires et que la troisième solution donnera un temps négatif, de sorte qu'il ne restera qu'une seule solution physiquement sensée.

[Franck Aurel]

Merci beaucoup j’avais pas pensé à séparer les equations

Answer 2

[David Goodmanson]

The syms solutions are getting out of hand, which is not all that unusual. The kinematic equations evidently are for a projectile lauched from location [-x0,y0] and striking the ground to the right at a specified [x,0]:

-x0 + v0*c*t = x                    % c = cos(theta)
 y0 + v0*s*t -(1/2)*g*t^2 = 0       % s = sin(theta)

The first equation is

-(x+x0) + v0*c*t = 0

The total specified displacement in the x direction is (x+x0) so renaming that quantity as a new x0 then

-x0 + v0*c*t = 0                    % c = cos(theta)
 y0 + v0*s*t -(1/2)*g*t^2 = 0       % s = sin(theta)

which describes a projectile lauched from location [-x0,y0] and striking the ground to the right at [0,0]. The total displacement is the only x quantity to worry about. That's another way of saying that since there are no external forces in the x direction, there is translational symmetry in x.

Instead of solving the second equation for t and plugging it into the first one, it's better to do it the other way round:

 t = x0/(v0*c)
 y0 + v0*(x0/v0*c) - (1/2)*g*(x0/(v0*c))^2 = 0
 or
 y0 + x0*(s/c) -(1/2)*(g/v0^2)*x0^2*(1/c^2) = 0
 define the distance
 b = 2*v0^2/g
 go to s/c = tg             % tan(theta)
 and use the identity (1/c^2) = (1+tg^2) to obtain
 y0 + x0*tg - (x0^2/b)*(1+tg^2) = 0
 or
 y0-(x0^2/b) + x0*tg - (x0^2/b)*tg^2 = 0 

then find the roots of this quadratic in tan(theta) and proceed from there.

There are two solutions, a 'slow' one where the projectile arcs up and takes a while to arrive, and a 'fast' one where the projectile is pointed below the local horizon at y0.

x = 0;                                    % for example
x0 = 0.08 + x;                            % new x0 is total x displacement
y0 = 0.05;
v0 = 2;
g = 9.81;
b = 2*v0^2/g
tg = roots([x0^2/b,-x0,x0^2/b-y0])          % two values
theta = atand(tg)                           % result in degrees

     b = 0.8155

     tg =
       10.6958
       -0.5022

     theta =
           84.6587
          -26.6641

% check
t = x0./(v0*cosd(theta))
y0 + v0*sind(theta).*t - (1/2)*g*t.^2      % should be zero

      t =
        0.4297
        0.0448

      ans =
        1.0e-15 *
        0.2220
       -0.0017