It should be hitting high, but a foot at 50 yards seems too high. Are you sure it's actually zeroed at 300?
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In my head I imagine this as two lines intersecting at a 45° angle at 300 yards (yeah I know thats an exaggerated example) with the top line representing the sight picture and the bottom line representing the travel of the bullet.
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The sight line will be a straight line from your eye, through the sights to the target.
Because the bullet is going to be pulled down by gravity as soon as it leaves the bore, the sights are set to angle the barrel upwards slightly with respect to the sight line. That will lob the bullet upwards, above the sight line, on an arc that will allow gravity to bring it back down so that it coincides with the sightline at the zero distance.
So the bullet is going to start out a little below the sight line (since the bore is below the sights), rise above the sightline (probably somewhere around 25 yards from the muzzle) and keep arcing upwards until the point where it reaches the peak and starts to descend.
The peak of the trajectory will be more than halfway to the zero point. In a vacuum, the high point of the trajectory would be exactly halfway to the zero point--the trajectory would be a perfect parabola. But because of air resistance, the bullet is slowing as it goes downrange and the parabola gets skewed. The bullet has to be fired so that it will reach the high point after the halfway point to make everything work.
Anyway, after the trajectory peaks, from that point on, the bullet will be falling. It will fall until it comes back down to the sight line at the distance for which the sights have been zeroed. It will keep falling, faster and faster the farther it goes downrange since the acceleration of gravity is constant and constant acceleration means constantly increasing speed.
This website has a good picture showing the relationship of the sights and the bore.
http://www.mainviewtech.com/en/branded.php?pid=193