I love this picture: The wave height would have to increase so dramatically.
The zoomed in image you present shows waves colluding and combining past eye level to obscure 1/4th of the distant Toronto skyline.
As a rebuttal, lets check this image against other sea-level images of Toronto from across Lake Ontario taken on days with different weather and wave conditions:

This image was taken from
http://www.weatherandsky.com/Mirages/Mirages.html, where the photographer describes it as a "View of Toronto Skyline (53km across the lake) through Canon Rebel digital camera 28 to 400mm zoom."
You should note that Toronto is not broken by the horizon line. How does Round Earth Theory explain this?
This particular image is devastating to the globe theory, because at 30 miles across Lake Ontario the earth should drop nearly 600 feet (60 stories). Quite clearly, we can see objects at the bottom of the Toronto skyline that are
not 600 feet in hight.
Here's the math:

Suppose that the earth is a sphere with a
radius of 3,963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangle as in the diagram.
Looking over a distance of 1 mile, we can use the theorem of Pythagoras:
a
2 = 3,963
2 + 1
2 = 15,705,370
and when we square root that figure we get a = 3,963.000126 miles
Thus your position is 3,963.000126 - 3,963 = 0.000126 miles above the surface of the earth.
0.000126 miles = 12 in * 5,280 ft * 0.000126 mi = 7.98 inches
Hence after one mile the earth drops approximately
8 inches.
-
Ergo, looking across 30 miles the Pythagorean theorem becomes:
a
2 = 3963
2 +30
2 = 15,706,269
and when we square root that figure we get a = 3,963.113549 miles
Thus your position is 3,963.113549 - 3,963 = 0.113549 miles above the surface of the earth
0.113549 miles = 5,280 ft * 0.113549 mi = 599.53872 feet
Hence after 30 miles the earth drops approximately
600 feet.
-
See
Chapter 2 of Earth Not a Globe by Dr. Samuel Birley Robowtham for a handy chart.