It is difficult to apply manually the same force and perform the same work for each throw. There could be some physiological reason for what you observed, it should be tested quantitatively. It is also possible that the angle at which you release the object has unwanted changes depending on the weight.
However, what you experience could be tentatively explained in the following simplified way.
For progressively more heavy objects, probably you reach a weight above which the performed work is approximately constant, say $W$. This is likely what you obtain trying to apply the same force along the distance in which the object is hold by the hand. $W$ is accumulated as kinetic energy $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the speed at which the object is released by the hand. So, roughly, $v=\sqrt{\frac{2W}{m}}$. The heavier is the object, the smaller is the speed $v$. Increasing even more the weight probably will decrease $W$ (but this is a matter about physiology) and thus $v$ will be even smaller.
In the opposite limit, for light objects, their weight does not matter too much: the speed $v$ is simply the final speed of the hand, when it releases the object. The movement of the arm is not affected too much by the presence of a light object hold in the hand. Thus $v$ reaches a constant value $v_0$.
Now we must discuss the relation between $v$ and the distance at which the object is thrown (at which it falls to the ground), with the assumption that the starting angle is the same. Here the friction of air comes into play. According to Stokes's law (see Wikipedia), the force is proportional to the radius $R$ of the object, $F\propto R$. In turn, the deceleration is $\frac{F}{m}\propto \frac{R}{m}$. If we assume objects with similar density, $m\propto R^3$, thus the deceleration is $\propto \frac{1}{R^2}$. Smaller objects are decelerated more than big objects, at least if their shape is almost spherical and their density is similar.
In conclusion. Progressively more heavy objects tend to make parabolic trajectories not affected by air, but their $v$ decreases with increasing $m$. Progressively more light objects tend to start with the same $v=v_0$, but they are more and more decelerated by air as $m$ decreases. The optimum is in between.
A last comment. The effect of the dimension of the object on air friction can be seen in the formula for the terminal velocity of a sphere, i.e. the velocity approached by the sphere falling for a long time in a fluid. This final velocity is $\propto R^2$. Bigger objects tend to fall faster than smaller object. A human body falling from an airplane reaches a speed of 190 km/h and needs a parachute to survive (see Wikipedia); it reaches the speed in 12 s, corresponding to a fall of roughly 300 m. A small spider falling from the roof of a building reaches its terminal speed of (say) 5 mm/s almost instantaneously, slowly falls along the whole building and reaches the ground without harm. Very small particles, with diameter less than 1 $\mu$m, can stay suspended in the slight turbulence of a gentle breeze permanently.
Best Answer
Interestingly, the only thing that matters is the shape (or distribution of mass) of the ball. Consider the total energy for an object rolling down an incline:
$$E = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2 + mgy$$
Now note that the rotational inertia of any rolling object can be written in the form $I = \beta mr^2$, where $\beta$ is some unites constant. For a solid sphere, $\beta = 2/5$, for a hollow sphere, $\beta = 2/3$, for a hollow cylinder, $\beta = 1$, while for a solid cylinder, $\beta = 1/2$. You can look these up in any physics textbook, or calculate them by performing an integral.
Also note that the "rolling condition" means that the angular speed is fixed by the speed of the center of mass and the radius: $\omega = v/r$.
Making these two substitutions, we can write:
$$E = \frac{1}{2}(1+\beta)mv^2 + mgy$$
Now imagine that we start from rest at the top at some initial height $y=h$, and roll down to $y=0$ at some final time. Imposing conservation of energy:
$$mgh = \frac{1}{2}(1+\beta)mv_f^2$$ $$v_f = \sqrt{\frac{2gh}{1+\beta}}$$
In other words, the final velocity is independent of the mass and the radius! It only depends on the factor $\beta$ which is determined by the shape of the rolling object (and the initial height, of course)!
EDIT:
I thought I'd add this edit about the acceleration. One can determine the acceleration also by conservation of energy. Start with:
$$E = \frac{1}{2}(1+\beta)mv^2 + mgy$$
and take $\frac{dE}{dt}$. Be sure to impose the chain rule since $y=y(t)$, and $v=v(t)$. Then, conservation of energy implies
$$ \frac{dE}{dt}=(1+\beta)mv \frac{dv}{dt} + mg \frac{dy}{dt} = 0$$. Now $y$ is the height, which we can express in terms of the distance $\ell$ along the incline (from the bottom) as $y = \ell \sin \theta$. But, $\frac{dv}{dt}=a$ and $\frac{dy}{dt}=\frac{d\ell \sin \theta}{dt}=v\sin{\theta}$, assuming $\theta$ is the constant angle of the incline and $v=\frac{d\ell}{dt}$.
$$ \frac{dE}{dt}=(1+\beta)mv a + mg v \sin\theta = 0$$. So the acceleration is: $$a=\frac{-g\sin\theta}{1+\beta}$$ Note that it reduces to the acceleration $a=-g\sin\theta$ of a sliding mass when $\beta \rightarrow 0$, and to "free fall" when $\beta \rightarrow 0$ and $\theta \rightarrow \pi/2$.