Say the ball's center is moving along a time trajectory that can be described as x = a t + b
and y = c t + d
-- any linear, uniform-speed motion can be described this way. Since you say that it's initially (say at t=0) outside the rectangle, we know that at that time x < 100 or x > 200, or y < 0 or y > 50 (one of the conditions of x, and one of the conditions of y, can both be true, but at least one must be -- if they were all false we'd be inside the rectangle).
So check "at what time and exactly where will that point intersect each of the four lines that make up the rectangle"; i.e., solve for t when x = 100 (which gives t = (100 - b) / a
, and therefore y = c (100 - b) / a + d
), x = 200, y = 0, y = 50. Discard the solutions where t < 0
(those were things that happened in the past), as well as ones where the other variable falls outside of the rectangle's boundaries (for example, for the t = 100 case I just mentioned, you can ignore the apparent solution if (100 - b) / a < 0
, or c (100 - b) / a + v < 0
, or c (100 - b) / a + v > 50
). If none of the four is left, this means the ball (with a radius of 0...) will not hit the rectangle along its current trajectory (it may if and when it bounces and thus changes trajectory, but those will be separate computations). If one or more are left, the one with the minimum value of t
is the one you want. Once you know where and when the center would hit the rectangle, taking account of the radius can be done separately, but won't change the issue of which rectangle side the ball hits.
The cases where the ball "glances" (hits the rectangle just because it does have a radius greater than zero) are harder, but one approach is, if the normal computation shows the ball "not hitting", repeat it after shifting the ball (by the amount of its radius) to both side of the trajectory-line it's following -- this will tell you if the ball IS in fact going to hit, and, if so, which side (assuming hits on corners can be counted as hits on one of the sides converging on that corner;-).