I''m trying to calculate the launch angle for a mortar-like weapon for an RTS game I''m working on. In the 2D case: Given a target (x, y) a launch velocity v and gravity g, what is the angle required to shoot a projectile from (0, 0) to hit the target? I worked out the solution for hitting a target at (x, 0): 2*angle = asin((g*x) / (v^2)) but I''m stumped as to the solution for the general case. Any help would be appreciated.

yet again another physics question i think the following equations are correct

@ = theta is the angle
V = velocity
g = gravity
t = time
Vy and Vx are velocities in x and y direction

horizontal velocity component

Vx= Vx0= V cos @ = constant

vertical velocity component

Vy = Vy - g*t = V sin @ -gt

horizontal position component

x = Vx*t = (V cos @)t

vertical position component

y = Vy*t - .5gt^2 = (V sin @)t - .5gt^2

Those equations are correcty but you seem to have missed the point of my post. I''m trying to find theta given a target (x, y) and a launch velocity. What you''ve posted gives the velocity components for a known angle.

If the velocity remains constant, it will look quite funny when the target is close to the tank.. The tank will shoot *very* high shots . I''m not sure if you really want that.

I suggest using constant angle (eg. 45'') and a varying velocity. And it''s very easy to implement too since the flying distance is:

R=(v0^2*sin (2*angle))/g

Solving v0 leads to

v0=sqr(R*g/sin (2*angle))

Check over my algebra before you use my final answer exactly... sort of odd to do this sort of thing without pencil and paper. Could lead to careless mistakes .

Otherwise, this is the kind of logic you need to use, along with a trigonometric identity, and finally the quadratic equation to separate the two possible firing solution angles. (Which happen to converge in some special cases.)

given: (0,0) -> (x,y) && v, -g | find: @_1, @_2

V_x = V*cos(@)
x = V_x*dt
dt = x/V_x = (x)/(V*cos(@))

V_yi = V*sin(@)
V_y = V_yi -g*dt = V*sin(@) -g*dt
y = V*sin(@)*dt -.5*g*(dt)^2
y = V*sin(@)*[(x)/(V*cos(@))] -.5*g*[(x)/(V*cos(@))]^2
y = x*tan(@) -(.5*g*x)/(V*cos(@))^2

identity:
tan^2(@) + 1 = sec^2(@)

0 = x*tan(@) -(.5*g*x*tan^2(@))/(V) -(.5*g*x)/(V) -y
tan(@) = (-x +- sqrt(x^2 +((g^2*x^2)/V^2) +(2*g*x*y)/V)/((-g*x)/V)

@ = tan^-1[(-x +- sqrt(x^2 +((g^2*x^2)/V^2) +(2*g*x*y)/V)/((-g*x)/V)]

therefore:
@_1 = tan^-1[(-x + sqrt(x^2 +((g^2*x^2)/V^2) +(2*g*x*y)/V)/((-g*x)/V)]
@_2 = tan^-1[(-x - sqrt(x^2 +((g^2*x^2)/V^2) +(2*g*x*y)/V)/((-g*x)/V)]

Well,

I''m not quite sure on how this game works, ie is it overhead? If your trying to hit a target on a plain, I would work in vectors. You seem to have the x direction worked out, all I would do is use the same formula, switch y for x and that would be my answer with respect to y. Then add the x and y vectors together, that will give you location.

well basically i forgot this part but the angle is this

tan @ = Vy / Vx