Does a long vertical pole fall at a different speed than a short vertical pole? The formula for a falling object has $r^2$ in the denominator.  This would mean that an object that is higher up falls more slowly than the standard $9.807\ \mathrm{m/s^2}$ that we are taught in high school.
What would happen if we took at $1$ metre pole and a $10$ metre pole up to a height of $100$ metres for the bottom of both poles, and then dropped them?   Let's assume they are weighted on the bottom so both remain vertical, and that the $10$ metre pole is hollow so they both weigh the same.  Would they hit the ground simultaneously or otherwise?
 A: Assuming (reasonably) that both objects are rigid (undergo no deformation under the influence of gravity) then in the Law:
$$F=G\frac{mM}{r^2}$$
$r$ refers to the distance between the centre of gravity (CoG) of the object and the CoG of the Earth.
Then assuming the CoGs of both objects have the same mass and are at the same height above the Earth (at the start of their free fall), the resulting $F$ are equal.

Would they hit the ground simultaneously or otherwise?

In those circumstances the former would be true.
A: Yes, slightly different acceleration as others have said, leading to slightly later arrival of the longer pole. (It is the location of the centre of mass which determines the acceleration).
An interesting related effect: the poles are both drawn into tension owing to the different gravitational pull on their two ends. The extreme example of this is the 'spaghettification' predicted to happen as things fall towards the centre of a black hole.
A: Ok, no air. Also I simplify out the tidal effects (poles are in no sense spherical). Mutual attraction between poles, as well as the Earth moving towards them assumed negligible (not that it will change the answer).
Other than that, 10m pole has its centre of mass 4.5m higher so it gets somewhat less gravity and less acceleration.
Since both poles have to travel 100m (equal distance), the 10m pole will arrive later.
A: Yes.  As described in the questions, there is very small difference between the acceleration of the two poles with the shorter one accelerating faster.
The difference
The gravitational force acting on a pole is
$$F = \frac{GMm}{r^2},$$
where $M$ is the mass of the Earth, $m$ is the mass of the pole, and $r$ is the separation between the center of mass (CoM) of the Earth and the CoM of the pole.  Neglecting air resistance and the gravitational effect of one pole on the other, the acceleration of a pole is
$$a = F/m = \frac{GM}{r^2}.$$
The masses of the poles don't matter.  They could be different or the same.
If the two poles are different lengths, $L > \ell$, then their CoMs will be at different distances from the CoM of the earth.
Let's define $R$ as the distance from the CoM of the Earth to the bottom of the polls, which are at the same height.
Assuming the poles are uniform
$$r_\ell = R + \ell/2 \quad\text{and}\quad r_L= R + L/2.$$
The shorter pole will will experience a larger acceleration.
$$ a_\ell = \frac{GM}{(R + \ell/2)^2} \quad > \quad a_L = \frac{GM}{(R + L/2)^2}$$
How big is the difference?
To get a handle on how much larger, we can do a first order expansion of the two accelerations and look at the difference.
$$a_L = \frac{GM}{r^2} = \frac{GM}{(R + L/2)^2} = \frac{GM}{R^2(1+\frac{L}{2R})^2} = \frac{GM}{R^2}\left(1 + \frac{L}{2R}\right)^{-2}\approx \frac{GM}{R^2}\left(1 - 2 \frac{L}{2R}\right)$$
I personally find the fractional difference $\frac{\Delta a}{a}$ to be more illuminating than the absolute. So lets look at that by dividing out the common $GM/R^2$ bit.
$$\frac{\Delta a}{a} \approx \frac{a_\ell - a_L}{GM/R^2} \approx (1- \ell/R) - (1-L/R) = \frac{L-\ell}{R}$$
The radius of the Earth is about $6\times 10^6$ m, so we're looking at parts-per-million differences in the accelerations of the two poles.
A: Paul T. provides a good answer regarding the case where the height of the bottoms of the poles are the same (which is what was asked for in the question).  The main difference in that case is due to the different heights of the centers of mass of the two rods.
However, you might ask, what if the centers of mass of the two rods were at the same height, would there still be a difference?  It turns out that there will be, although the difference is even smaller.
Let $m$ and $l$ be the mass and length of a vertical rod of uniform density, and $r$ the height of its center from the center of the planet.  The planet's mass is $M$.  Consider a tiny piece of the rod, of mass $\delta m$ and distance $x$ from the rod's center (so $x$ is between $-l/2$ and $+l/2$).  The force of gravity on the tiny piece is:
$$\delta F=\frac{GM\delta m}{(r+x)^2}$$
The total force on the rod is the integral of $\delta F$ over the whole mass:
$$F=\int\frac{GM}{(r+x)^2}dm$$
The mass of the small piece is proportional to its length ($m/l=\delta m/\delta x$) so we can substitute $dx$ for $dm$ with the appropriate scaling:
$$F=\int_{-l/2}^{+l/2}\frac{GM}{(r+x)^2}\left(\frac{m}{l}dx\right)$$
Doing the integral yields:
$$\begin{align}
F&=-\frac{GMm}{l}\left.\frac{1}{r+x}\right|_{x=-l/2}^{+l/2} \\
&=-\frac{GMm}{l}\left(\frac{1}{r+l/2}-\frac{1}{r-l/2}\right) \\
&=-\frac{GMm}{l}\left(\frac{(r-l/2)-(r+l/2)}{(r+l/2)(r-l/2)}\right) \\
&=-\frac{GMm}{l}\left(\frac{-l}{r^2-(l/2)^2}\right) \\
&= \frac{GMm}{r^2-(l/2)^2}
\end{align}$$
This is almost the same value as if the mass of the rod was concentrated at its center (in which case it would be just $GMm/r^2$).  Like Paul T., let's look at the relative difference:
$$
\frac{\frac{GMm}{r^2-(l/2)^2}-\frac{GMm}{r^2}}{\frac{GMm}{r^2}} 
= \frac{(r^2)-(r^2-(l/2)^2)}{r^2-(l/2)^2} 
= \frac{(l/2)^2}{r^2-(l/2)^2} 
\approx \left(\frac{l}{2r}\right)^2
$$
Compare this to the case where we measure $r$ from the end of the pole, where the relative difference (between a rod and a point) was just $l/r$.  If the end case had a difference of on part per million, the center case will have a difference of less than one part per trillion!
A: In non uniform gravity, discounting air resistance,  any two objects having different heights for their centers of gravity will be accelerated by gravity at different rates. As you seem to understand, the increased radius to the center of gravity would give a lower rate of initial acceleration to the higher object..
A: Actually the question is a little vague. When I read it, I thought it meant like a tree, or a telegraph pole, or one of those very tall old chimneys falling over. In which case we are calculating the time for a tall thin cylinder to fall over. The taller it is, the longer it will take to fall although in real life we would have to contend with the cylinder breaking or deforming.
Most of the answers here focus on dropping a vertical cylinder and have answered that thoroughly. There is also the possibility of the cylinder (pole) being held horizontally before being dropped in which case the length would not be relevant.
I hope that has muddied the waters a little...
