如何为射弹获得一个角度，使其在分解重力的同时击中矢量2目标

Question

I have an enemy that can shoot projectiles, but my issue is I am trying to find the angle required to hit the player, but the projectile has to be shot at an exact velocity (ie can't shoot the projectile more slowly). I have tried the range equation, but I can not simplify it such that I get the theta (angle), that factors in the y distance, the x distance, the velocity, and the gravity.

Sorry for the sprawling question, Teh Cosmic Sloth

Answer 1

Well... You have: D - distance to target

g = 9.8 - g- force

v - your firing velocity

Your velocity along X is v*cos(a)

Your velocity along Y is v*sin(a)

you need to travel distance D, hence your time of flight is t = D/v*cos(a)

in that time you need to reach the maximum height of your flight and fall back. At maximum point your vertical speed will be 0, at the end of flight your vertical speed will equal to the starting vertical speed, if we ignore the air resistance.

So, bearing in mind that the equation for speed is v = acc*t (where acc is g for you) t = v/acc. In your case you have to drop speed to 0 and get it back to original, so you have factor of 2.

t = 2*(v*sin(a)/g)

Lets sum it up:

t = D/v*cos(a)

t = 2*(v*sin(a)/g)

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2*(v sin(a)/g) = D/v cos(a)

You know everything but angle a.

So, by doing some variable transfers you get

2sin(a) cos(a) = D g / v^2

which is:

sin(2a) = D*g/v^2

2a = arcsin( D*g/ v^2)

So, this must be the answer.

a = arcsin( D * g / (v^2) ) / 2