Paradox- Object thrown parallel to the ground will never fall down Suppose an object is thrown parallel to the ground. The gravity acts downward (ie. perpendicular to the direction of motion of the object). The work done by gravity on that object will be given by : 
$\text{Work}, W = F \cdot S = FS\times\cos \theta$
$\implies W=FS \cos 90$ ($\because$ direction of gravity $\perp$ direction velocity of the object)
$\implies W= FS \times 0$ ($\because \cos 90 = 0$)
$\implies W = 0$
From the calculation above, gravity will not do any work on the object. So why does it follows a parabolic path and eventually falls down.
 A: Recall that a force perpendicular to the direction of motion does no work but simply changes the direction of the velocity vector.
The same thing happens here: Initially the ball's motion is perpendicular to the force of gravity and hence at this very moment, gravity does no work but slightly "rotates" this velocity vector towards the downward direction; as soon as this velocity vector is a little rotated, it is no longer perpendicular to the force of gravity and thereafter gravity starts doing work on the ball. In your example, gravity does no work on the ball at the very instant it is thrown but starts doing work after that.
A: Your basic assumption is that there is no component of force in the downward direction. Assuming this is true, then your calculations are right.
But the force on a body $F$ is given by $$F = ma$$ where $a$ is the acceleration and $m$ is the mass of the body. In the case of gravity, there is an acceleration due to gravity that acts on the body. The acceleration is in the downward direction, and hence there is a component of force acting in the downward direction.
Therefore the correct way to solve this problem will be to use vector resolution, and resolve the force into two components and solve for the dynamics along each component.
A: Your calculation is incorrect.
$\text{Work} = \text{Force} \cdot \text{displacement} = F \cdot s$
The above product is a "dot" or "scalar" product, which means we only consider the displacement that occurs in the direction of the Force, which in the case of gravity is downwards.  Can we set this vertical displacement to 0?  No we cannot, and here is why:
$\text{Force} = \text{mass}\times \text{acceleration} = mg$, where $g$ is the acceleration due to gravity.  Therefore the object being thrown will accelerate downwards with an acceleration of $g = 9.8\,ms^{-2}$.  
We then can use the equation:
$s = ut + \frac{1}{2}at^2$.  In the case that the object was thrown horizontally, $u = 0$, therefore, $s = \frac{1}{2}gt^2$, where $s$ is the displacement in the vertical direction.
Therefore, even though the object is thrown horizontally, there will be a non-zero displacement in the vertical direction, and therefore there will be a non-zero value for work.
A: Along the vertical line of the place where the throwing happens, the object does not have velocity, but it is at rest. Since the only force which acts upon the object is its weight - a vertical and downward force that translates interaction between any object and the earth -, the object will at leas present a downward motion with acceleration along that vertical of the place, but without any acceleration along the parallel of the place. It means the object develops a motion which is compounded by a horizonal motion (the velocity does not vary) and a vertical motion where the velocity varies. These considerations show, therefore, that the object will eventually touch the ground.  
