PPT Slide
Well, there are two nuances here and they both involve the vertical height of the reflection in the water.
First, the height is always judged from the reflection plane and not the base of the object itself. In the diagram, the end point ep of the pole is not reflected around its station point sp but around its station point in the plane of the water surface, spW.
For objects standing along the edge of a gently sloping beach or marsh this is not a distraction, because the ground level is roughly the same as the water level. However the diagram shows the pole standing on a raised embankment, and this distance has to be included in the total length that is included to find the reflected end point, epR.
The second nuance is that the object and its reflection will have an identical height only if the direction of view is nearly parallel to the water surface. This is the case in those dramatic postcard views of huge mountains seen reflected in the surface of an enormous lake. In the example above, however, the reflection of the pole is clearly viewed at a downward angle, which will make it appear foreshortened in comparison to the actual pole.
No problem: because the perpendiculars are parallel to the image plane, their measure points are the diagonal vanishing points or distance points (dvp) at the top or bottom of the circle of view. Just use these diagonals to determine the correct height of the reflection, using the methods described here for projecting a measure bar or unit dimension into perspective space.
That is, (1) draw a line from spW to any convenient point along the horizon line (the example uses vp1, but any point will do), (2) project in depth the vertical distance to be reflected with a line from ep to a dvp to define point R1, (3) draw a vertical line through R1, (4) get the location of R2 by a diagonal from spW to the same dvp through the vertical line, and (5) locate epR by a line from your vanishing point (vp1) through R2 to a vertical line under spW. Piece o' cake