Understanding the trajectory of a projectile It is stated that a projectile's motion can be divided into individual vectors. Now according to the laws of physics, the forces acting in the y-direction will never cancel or affect the force acting in the x-direction. Hence the weight $mg$ and the horizontal force $F$ are independent of one another. 
If that is the case, then why does a ball take longer to reach the ground if the horizontal force acting on it is large? 
Imagine it this way, a ball machine shoots a ball with a great force in the x direction, then why is it that the ball takes longer to fall when the x component is so large? Shouldn't the force $mg$ cause a constant downward acceleration?
I understand that in case the x component is very large, the distance covered will naturally be greater but I'm speaking in terms of time taken for it to reach the ground.
 A: If you consider the Earth to be flat, adding horizontal velocity (or force in your example) won't change how long it takes for the object to fall.
It covers more distance in the x direction in the same amount of time, because it has a higher velocity.
The time to fall depends only on gravity, because only gravity will make it fall.
If you have a very high horizontal velocity it will change the time it takes to fall, because then the Earth can't be approximated as flat over the distance you travel. As you move horizontally the Earth will curve further below you.  If you can balance the rate of falling with how fast you are going, you will orbit the planet instead of falling (which is what we do with satellites).  It's kind of like you're constantly falling forward so fast that you never reach the ground.
Note: This answer is assuming we are not considering drag (which may or may not be appropriate, it depends entirely on how basic you are assuming the system is.
A: For a real projectile, there are two forces at work during the flight: gravity, and drag. Now drag is a quadratic force - that is, when you double the velocity, the force becomes four times greater:
$$F = \frac12 \rho v^2 A C_D$$
In this equation, $\rho$ is the density of the medium (air), $A$ is the projected area (cross section) of the object, $v$ is the velocity, and $C_D$ is the drag coefficient (a function of shape, and of Reynold's number). For a sphere we usually assume $C_D = 0.5$ but that is an approximation.
Now let's draw a diagram of a projectile in flight, having just "dipped" away from the horizontal direction. I draw the diagram for two different horizontal velocities, and compute the vertical component of the drag.

As you can see, the larger horizontal velocity gives rise to a larger vertical drag component - so if quadratic drag is present and non-negligible, the projectile will indeed fly further, and stay airborne longer.
Surprising, isn't it?
What about curvature
If you ignore drag, but include curvature of the earth, then the argument goes like this: if you shoot a projectile horizontally from height $h$ so it would normally land at a distance $D$ (on a perfectly horizontal surfaces), then the earth will have "curved away" a little bit in the meantime. How much? For small distances, we can calculate the "dip" $d$ as
$$\begin{align}d&=R(1-\cos\theta) \\
&\approx R\left(1-\cos\left(\frac{D}{R}\right)\right)\\
&\approx \frac{D^2}{2R}\end{align}$$
Where $\theta$ is the angle subtended between the start and end of the trajectory, seen from the center of the earth. When you shoot an object so it lands 100 m away, the curvature would add an additional 0.1 mm - negligible. Shoot 1 km, and it becomes 8 cm - still very little. Shoot 10 km, and the "dip" is 7.8 m, it would have a measurable effect on the time to drop. But compared to the drag effect, it is still very small.
A: 
I'm trying to explain without mathematical calculations.

First of all, the horizontal and vertical components work independently.
Secondly, the usual treatment is assuming the magnitude of the initial velocity is fixed.
Thirdly, the time of flights (assuming you projected the object from horizontal ground and it's landing on the ground at the same level) is the time requires to return at the same level.
So larger the angle of projection (from horizontal) means larger the vertical component of velocity; as a result, longer the time of flight.
A: Your model is not taking in consideration the resistance of the air, which is a fluid. That resistance adds a new force on the Y-axis which compensates the weight.
In a vacuum, the ball would reach the ground at the same time, for any X-axis speed.
