accelerationdragnewtonian-gravitynewtonian-mechanicsprojectile
Terminal Velocity depends on two things: surface area and speed. These are inversely proportionate.
If both these variables affect terminal velocity, why do parachutes slow you down? Initially you had a small surface area but a fast speed- with the parachute you have a larger surface area but lower speed. You have increased one variable but decreased the other. Therefore why do parachutes decrease speed?
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.
There are a couple issues
EDIT: to get a feeling for 1. we can first calculate the Reynolds Number using this site
Since the flow is most likely turbulent the friction is quadratic, i.e. $$ma=-C v^2+mg.$$ Using $v(0)=0$ and a little help of Mathematica we can solve this to get $$v(t)=v_t\tanh\left(\frac{t}\tau\right)$$ where I define $\tau=\sqrt{\frac{m}{gC}}$ as the characteristic time and $v_t=g\tau=\sqrt{\frac{mg}C}$ as the terminal velocity. Let's now plot this function
From this plot we can conclude that if the time is greater than about 2-3 characteristic times then the object has reached terminal velocity.
Best Answer
$(Drag) = (Coefficient)\times(Area)\times(Speed)^2 $
When $(Drag)=(Weight)$ then
$$ (Speed) = \sqrt{ \frac{ (Weight) } { (Coefficient)\times(Area) } } $$
So the larger the area the less the speed. What is the question again?