True, but you could just see the air resistance as 2 separate orthogonal components, one in the oppositie direction of gravity and one peripendicular to that vector. One will behave as I described while the other will just result in the the melon slowing down its horizontal movement.
Without air resistance, assuming the launch velocity is the same, the arcs should be the same, as the only force acting will be gravity, resulting in a constant downwards acceleration regardless of mass.
With air resistance the arcs would deform a bit, as during the initial phase when the melon is going up the air resistance’s vertical component would be aiding gravity, whereas in the second phase it would be opposing it. Horizontal just lowers horizontal speed over time, although depending on vertical acceleration this might affect the shape a bit.
I phrased that a bit poorly, if the launch velocity is the same the curve will be exactly the same, so the distance will even be equal. The size just scales in launch velocity, the shape is indeed the same.
Depends a bit on the shapes of the objects involved. Since air resistance is based on both the shape, surface area and current velocity of the object, changing one of these will change the force, deforming the curve. I assume arrows and melons have different surface areas and drag coefficients, so their exact arcs won’t be the same.
With air resistance, even if the heavy and light objects have exactly the same shape and size, their paths will not be parabolas and they will not have the same shape as each other. The paths they trace out will not be relatable to each other by any method that involves any combination of moving, stretching, or shrinking their paths, whether vertically, horizontally, or both, even if we allow different amounts of moving/stretching/shrinking in different directions.
The reason why this is true is that unlike gravity, air resistance does not provide constant acceleration. Only launched objects subject to constant acceleration travel in parabolas.
Here is one way of showing what I said is true. First, we argue that they won’t be parabolas. To see this, note that if I have any object which travels in a parabola, its position as a function of time is quadratic. If we differentiate this twice to get acceleration, we get a constant.
Now to see that the shapes of the arcs of the heavy and light objects won’t be the same, note that a very heavy object will not be affected much by air resistance but a light object will. Thus, if we consider launching objects of various masses, as we consider objects whose masses get higher and higher, the path traveled gets infinitely close to a parabola. On the other hand, the path of a lighter object will have at least some finite difference from a parabola.
 Technically, even gravity does not provide constant acceleration. If you are closer to the earth or any other spherical planet, you will experience more acceleration from gravity, although this difference will typically be a lot smaller than air resistance. Also, some parts of the earth have slightly higher gravity than others due to differences of density of the earth and the fact that the earth isn’t a perfect sphere. Once again, this difference is small.
The drag due to air resistance is proportional to the velocity of the projectile. Since it’s moving fastest immediately after launch, the drag is greater at the beginning than the end. A ballistic flight path under air resistance conditions looks something like this:
I decided to visualize the trajectory of the melons a bit:
I simulated throwing the melons with an initial velocity of 10m/s at a 45 degree angle. I varied the melon radius, and adjusted the mass accordingly by taking the average density of a watermelon and multiplying that by the volume of a sphere. I also used the drag coefficient of spheres, as I thought that would be close enough to approximate a melon. The three different bars in the graph show the melon radius in meters.
Assuming I did the simulation right (it’s 5AM, I really shouldn’t be writing physics simulations), the lighter melon is more heavily affected by the air resistance, giving the same starting velocity and angle. Now how you’d actually launch a 1m melon is an entirely different question.
The paragraph you referenced was arguing that there is no way of relating the paths of the light/heavy objects by moving/stretching/squashing them in any combination of directions. The reason is that if there were, then we could turn the lighter object’s path into something infinitely close to a parabola by moving/stretching/squashing it. However, the paragraph before the one you referenced argued you can’t do that.
 Technically, the earlier paragraph just argued the lighter path can’t exactly be transformable into a parabola by moving/stretching/squashing, but a more detailed argument can show that you can’t be infinitely close to being approximately transformable into a parabola either. Also, in the sequence of heavier and heavier objects one considers, I should probably have required they all be launched with the same velocity which is the same as that of the lighter object.
The amount of energy Epot stored in a drawn bow is dependent on properties of the bow itself, and is largely independent of the projectile.
Thus, if you launch a heavy object and a light object using the same bow, the heavy object will have a lower inital speed v0, related by the square root of the ratio between the masses.
This lowers the amount of horizontal distance it can cover during its time of flight, thereby making its descent steeper.
For a useful demonstration, consider launching a melon with inifinite mass. It will hit the ground at a perpendicular angle; the steepest descent possible.
On the other hand, consider launching arrows of a progressively lighter mass ε. In the limit as ε↘︎0, its initial speed will approach ∞. Assuming you shoot it parallel to the ground, it will hit the ground at an angle limε↘︎0 sin-1(sqrt(ghε/Epot)) = 0π – the least steep angle there is.
Edit: Math may be wrong; I’m sleepy and on my phone. The point is, it’s either 0 or ∞, which is all that matters.
I assume that the bow is at least some distance off the ground, in part because that is normally the case when using a bow, but also because otherwise the projectile will need to pass through the ground while the bow is drawn.
Unless you’re firing from a hole, but then the ground would not be uniformly flat, and that’s a bit too out there, even for a thought experiment.
That’s with the assumption that the launch velocity is held fixed. In reality, bows don’t launch objects with a fixed velocity, they launch with a given amount of force, so the heavier the melon, the lower the launch velocity, so the closer in front of you it’s going to land, so the more vertically it’s going to hit.