Let's assume that the bullet lodges in the target (which doesn't happen, it passes through, but my point will still remain). If that happens, we can analyze the situation using momentum.
Let:
M = mass of human
m = mass of bullet
v = muzzle velocity of bullet
V = velocity of human and bullet after impact
Because
[1] (M+m)V = mv
We can calculate V as follows:
[2] V = mv / (M+m)
Let's call an average target 180 pounds. A typical 50BMG cartridge (i.e., the U.S. M33), is 668 grain and has a muzzle velocity of 2910 fps. Converting everything to metric, we have the following values for our known variables:
M = 81.65 kg
m = 0.04329 kg
v = 887.0 m/s
Plugging these values into equation [2] above, we can determine that V = 0.4700 m/s, or 1.542 fps.
Analyzing these results, it appears that, if a person is struck by a 50BMG round at full velocity, and the bullet lodges completely within them so that the full force of the bullet is felt, we can expect the target to stumble back about 1.5 feet. If they drop, it will either be because of a psychological reaction to being shot, or because of physiological incapication resulting from a central nervous system hit.
In reality, the bullet will not lodge itself in the target, and the velocity of the bullet will decrease over distance, so these results will be reduced markedly. So, no, those targets cannot be human --- it would break the laws of newtonian physics, and we're not dealing with general and/or special relativity speeds here, so quantum physics do not apply. And it is absolutely impossible to injure a person by having a .50BMG whiz by --- it must hit them.
BTW, your average prairie dog, with a weight of 2.5 lbs = 1.134 kg, will be moving at 32.62 m/s = 107.0 fps after a perfectly inelastic collision with an M33 round at full velocity. Hence my opinion as to the species of the targets in the video.