? When an object passes through the same H in the process of free fall, the time is the same, so the time from V 1 to N, V2 to G is the same. Because it is an elastic collision, MG=GE=EN. Then the distance traveled at the same time is 3MG=MN, that is, 3V 1=V2.
? Now let's look at point P, which is the point where two strokes pass. Now, suppose that one ball reaches point N through V 1 and the other ball reaches point G from point F through velocity V2, and they are thrown horizontally at the same time, then it must also land at the same time, which is beyond doubt. Because they are carried out at the same time, the time from P to G at the speed of V2 must be equal to the time from P to N at the speed of V 1. Their time is equal, and the horizontal speed is 3V 1=V2. Then it can be said that 3GQ=QN and MQ=QN, because MG=GE=EN.
Ok, now just watch the ball go through the model from p to n at the speed of V 1 Because the projection point Q of point P is the midpoint of the trajectory projection in the horizontal direction, the time from the ball to point P and the time from point P to point N in the vertical direction are also clear at a glance. According to H= 1/2(gtt), the displacement ratio of the ball in the vertical direction with the boundary of point P can be calculated.
The answer is H=4/3h, is that clear?