Why is the time taken for something to fall proprtional to acceleration due to gravity? This is a continuation of this question.
I saw in a piece of writing (that I no longer seem to be able to aquire) on dimensional analysis that:
$$t \propto h^\frac{1}{2} g^\frac{-1}{2}.$$
How can $t$ be proportional to $g$ when $g$ is a constant? I ask this because as $t$ rises surely $g$ can neither rise nor fall, which makes the thought that it cannot be proportional to $g$ spring to mind immediately.
 A: $g$ is an acceleration.  
So consider another similar situation with an acceleration: suppose you have a car starting at rest but which has a steady acceleration $a$ and you want to know how much time $t$ it takes to travel a given distance $s$.  Exactly the same dimensional analysis would lead to  $$t \propto s^\frac{1}{2} \;a^{-\frac{1}{2}}.$$ 
The signs here at least are intuitively obvious: if you have to travel further then you expect the time to be longer, but if your acceleration is greater then you expect the time to be shorter.  
And then you come to dmckee's point: if gravity is less then (ignoring anything like air resistance) the time increases compared with what you are used to on Earth, as you can see in the slowness of the feather and hammer Moon drop: that video also demonstrates that if there is nothing other than gravity operating then mass does not affect time.  The Moon's gravitaional acceleration at the surface is about one sixth of the Earth's so using the dimensional analysis you can predict that the time for the hammer to fall the same distance is about two-and-a-half times as long as it would be on Earth.  
A: @OllyPrice  Not sure this is what you want, or even if it will work here. New to this???
Yes g is a constant, so can be dropped from the equation.
So t is proportional to the sq.root of h
or h is proportional to t squared
I.E.  Distance h,is proportional to the square of the elapsed fall time, t.
A: $g$ is not necessarily a constant if you consider it as "the gravitational acceleration at point A on Earth", and more so if you consider other planets. 
$g$ varies around the earth--since the distance from the center of the earth varies. $g$ at mount Everest is lesser than $g$ elsewhere.
Aside from that, $g$ on the moon is approximately one-sixth of $g$ on Earth. So $g$ can vary.
Anyway, one can include relevant dimension-ed constants while doing dimensional analysis. In fact, one has to do so. Otherwise, with dimensional analysis, you will get the wrong expression--since by multiplying/dividing by a power of the dimensioned constant (which you will have to do sooner or later to make it dependant on the constant), the dimensions of the result change. Aside from this, you may have a two-equations-three-variables moment.
A more intuitive reason for why we include dimensioned constants--you can imagine that they changed, and predict the result based on that. In most cases anyway, the constant is not truly a constant, like $g$. The only "true" constants are $G,c,\hbar,R$, and parameters of various bodies. And some other stuff I probably forgot.
A: $y={gt^2}/2$ so $t=(2y/g)^{1/2}$ so $t$ is proportional to $g^{-1/2}$.
$g$ is not constant. It falls off the higher you go above the surface of the earth.
