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.

  • $\begingroup$ I love this "paradox", learn from it! $\endgroup$ Jan 30, 2013 at 5:20
  • $\begingroup$ quick question for you; would you say that gravity does no work on a body falling from rest at a height above the ground? $\endgroup$ Nov 2, 2013 at 16:59

4 Answers 4


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.

  • 2
    $\begingroup$ Just what I was about to post; this answer is absolutely correct and well phrased! $\endgroup$ Nov 2, 2013 at 16:56

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.

  • $\begingroup$ (as an interesting exercise): How does that translate into orbital motion? Assuming the ball is thrown at an altitude whereby there is no atmospheric drag and at a sufficiently high speed to orbit the celestial body, is gravity then considered to still do no work (since the direction of force is always perpendicular to motion), or is it that the work done to accelerate the object downward is counteracted by the object being at orbital velocity? $\endgroup$ Jan 29, 2013 at 15:23
  • 1
    $\begingroup$ @Justinᚅᚔᚈᚄᚒᚔ Here one has to consider how the line joining the centre of earth and the object (gravity acts along this line) changes direction as the object moves around. For perfectly circular orbits, gravity does no work since it is always acting perpendicular to the object's motion. This is because as the object's velocity vector 'rotates' due to gravity, the object would have travelled such distance that gravity becomes perpendicular to the new rotated velocity vector as well. In case of elliptical orbits, gravity does do work since the object's motion isn't always perpendicular. $\endgroup$
    – Alraxite
    Jan 29, 2013 at 17:37

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.