As the name implies, the lotus on the water is 0. 5. When the wind comes, the horizontal displacement on the water surface is 2. The problem should be that in an ideal state, since the stem of the lotus is straight (not curved), the trajectory of the flower is obviously round, so let the water depth be X and (X+0.5) be the radius of this circle. After the flowers are blown away by the wind, the horizontal distance of the water surface is 2, so the square of x is equal to the square of (X+0.5) minus the square of 2, that is, 4, X=3.75, so the depth is 3.75 feet.