So the ball is moving in a linear motion so we can describe the line by the an equation of the form y = k * x + m
We can easily calulate k an m:
k = ball.xVel / ball.yVel
m = ball.yPos - k * ball.xPos
now we can solve the equation when x=AI_X_POS
y = k * AI_X_POS + m
You might also want to take the wall bounce into the calculation but that should not be too hard when you know the basics.
Lol. Obviously the OP was asking about wall bounces. He wanted to know how to make predictions about where the ball will end up once it reaches the opposing side.
Not how to calculate movement in a straight line.
The answer is that you must compute the linear trajectory and see if it hits before the opponent's side, if not, you have the intersection. if it does, you calculate one more iteration, if this hits the opponent's side, you're done. if not, you have the interval length, now you go "mod that length" and then you only have to compute from the last interval to the opponent's side.
it's very simple and can easily take place in the time it takes to render a single frame. if you want an unscoreable opponent, this is all you need to do.
having it angle the ball to maximize the difficulty of your return is a simple linear system on points scored against you vs. the terminal angle of the ball as it approaches your side, using the reverse of the above calculation to force it.
pong has always been a game of dumbing down the opponent enough that the game is fun to play.
