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Tetris / Dropping a ball (elementary question)

I’m having some trouble understanding how dropping a ball through a tetris shaped hole works.
The question is: “A tetris shaped hole has sides of length 32 and a top of length 28. (the rest of the board is the same length) how do I calculate the height of the gap in the middle of the board for which a ball can be dropped to the bottom of the board”.
Any help is greatly appreciated!

A:

A ball, when dropped, hits the lower left corner of the board. Because of the size of the hole, when the ball is a little further from the center of the board, it will exit at the lower left corner (assume that the wall is perpendicular to the board) and when it’s a little closer to the center, it will exit at a higher point on the left side.
The problem is to find the minimum distance for the ball to exit at the left side of the board (it can only go through the upper left quadrant of the board as seen from above).
Also, you can find the distance it takes to hit a point on the top of the wall by considering the problem from the top. The distances added together should add up to the width of the wall you’re asking about (ie: $32+28$=$60$). Since the ball can only exit through the upper left quadrant, you’re looking for the minimum distance between the ball and the right edge of the board. When you place that ball at a distance $x$ from the right edge, the ball will exit at $(32-x)$. It is easy to find where the derivative of this function is $0$:
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[(32-x)\right]=0\implies x=32$$
This means the ball will exit at the bottom center when placed at $32$ from the right edge. This seems obvious, but

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. Hi, There is a good article written on the usage of rectangles in autocad. Read it out from here. There is not any way to check whether the shape is defective or not. KK Auto Electric Products LLC.
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