Why do things that are far away seem smaller? As you see things that are far smaller, a funny question about this:
Imagine there are many people in a row (all are same height)
    A            Other Guy B             C             D
    |                        
you |  distance     |        distance        distance 
    | <-------->    |        <------->   |  <------->   
    |               |                    |             |
    |               |                    |             |
    ------------------------------------------------------  floor   

From your point of View you see D much smaller than you, then..
Remember, if this is an "effect" then D is not smaller than you, it's just far away
(and this is not earth curvature, because D is smaller in all sense not only in height)
so..
Imagine you have a laser pointer and want to point your partner's face to annoy him
so 
how do you target it? 
do you align the laser with the floor at the height of your own face believing in geometry? or do you point directly at the "little person" D ?
    A               B                    C             D
    |. . . . Laser path . . . . .  . >. . . . . . >. . . . . . >. . .  . >
    |  ^            |                           
you |  |parallel    |                    |
    |  |            |                    |             |     
    |  v            |                    |             |
    ------------------------------------------------------ floor   

Then you don't hit him, and the "visual effect" became so real that you start to think D is for Dwarf. 
There is something Wrong? Physics opinions are welcome!
I think the answer is that floor itself will seem not aligned eigther
so targeting the Dwarf you will be really aligned with floor
because floor itself will have perspective effect.
    A               B                    C             D
    |. . . . Laser path . . .   
    |               |        ''''''''''''......
you |               |                    |      ......>|
    |               |                    |      _______|___floor   
    |               |         ___________|------                        
    --------------------------

Anyway its weird
 A: I will kindly assume that your question isn't a hoax.
Objects that are further from the observing eyes look smaller because of geometric optics. If an object at distance $D$ from our eyes has size $S$, the rays from its endpoints will arrive to our eyes from the same angles as the rays from a smaller object of size $S/k$ whose distance is $D/k$, because of two simple similar triangles. The angle is essentially $S/D=(S/k)/(D/k)$.
A single eye may only detect the direction from which a light ray is coming - the light ray will create a point on the retina that only depends on the angle from which it is coming (let's neglect focusing now - focusing either by lens in each eye or by the relative position of both eyes is able to determine the distance of close enough objects), so it will think that the large object of size $S$ is as small as the smaller object of size $S/k$ just because the larger object is $k$ times further.
A random picture sufficient to explain what I mean, see below:

In the picture above, "A" is the eye, the line interval "BC" is the closer and smaller object, and the "DE" interval is the larger but more distant object. The light rays from points "C" and "E" arrive to the eye "A" from the same direction, and similarly for the light rays from "B" and "D", so the eye can't really distinguish the objects "BC" and "DE".
Cheers
LM
A: This is a good question, and what you are encountering is what is often called the inverse r-squared scaling law, or more simply, the inverse square law.  In this case, we can write a crude relationship:
$$Size_{apparent} \propto Size_{actual}\dfrac{1}{R^2}$$
If we think in terms of steradians, we can think in terms of the area of spheres we can take the actual two dimensional size of an object (say 1.5m high and 0.5m wide = 0.75 $m^2$); and if we say that object is 3m away, we would compare that area to the area of a sphere that has a radius of 3m.
$$4\pi{R^2} = 4\pi{3^2} = 36\pi$$
$$\dfrac{0.75}{36\pi} = \dfrac{1}{48\pi}steradians$$
If we take that same object and move it out to 30m we find that its apparent size shrinks much faster than its linear distance:
$$4\pi{R^2} = 4\pi{30^2} = 3600\pi$$
$$\dfrac{0.75}{3600\pi} = \dfrac{1}{4800\pi}steradians$$
This is the reason that an object appears very small as you move it further away from you.  
If you exclude curvature effects of the earth, the object that is further away from you because its apparent cross section is much smaller than if it were closer.
In answer to your question, barring other physical effects (particularly curvature of the earth effects), if you wanted to hit an object of equal height that was further away with a laser, you would want to aim as straight and parrallel to the ground as possible, and above the ground at an appropriate height.
