Driving into work, I started thinking about the arc of something being thrown and was puzzled about how gravity's affect is squared per second for falling bodies. Intuitively that implies the shape would be "like" a parabolic shape. But I'm curious if it truly is parabolic. Doing some googling, I've found some rudimentary explanations that remind me of explanations I received in algebra as a kid (such as http://entertainment.howstuffworks.com/physics-of-football1.htm)

My confusion is around the idea that gravity's effect on falling bodies (reference http://en.wikipedia.org/wiki/Equations_for_a_falling_body) is increased over time. If this is the case, then wouldn't this affect the arc in such a way that it's longer on the release end and curves downward faster at the end? This might not be as noticeable for short distances, but intuitively it could play an important role in longer distances.

In other words, my question is pretty simple, as soon as something is thrown (shot, projected, etc) is it considered a falling body. Either way, is the arc or path truly parabolic, or is the path elongated on the throwing end and curve downward faster at the end? If it is truly parabolic, can you please give a clean explanation as to why the effect of gravity over time doesn't apply? If my intuition is correct about it being elongated, can you please share a useful reference as well?

A couple of assumptions:

  1. What is being thrown is small and close to a large body. Like throwing a football 1,000 miles across to the earth's surface.
  2. Ignore air resistance or other factors for simplicity sake.

enter image description here


The Newton cannon illustration in @HariPrasad's answer shows us the flight path is elliptical not parabolic. It shows how modifying the initial vector's magnitude, when the angle is tangent to the earth effects the ellipse. It however does not show how changes to the initial vector's angle affects the ellipse.

Can we formulate an equation for the path (Reference: https://en.wikipedia.org/wiki/Ellipse)? I'm hoping for an answer that explains how the foci positions, and sum of red and blue line (need editor to give technical name) change in relation to changes in the direction and magnitude of the initial vector.

Animation of drawing an ellipse from wikipedia

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    $\begingroup$ Given your assumptions, yes, it is parabolic. Gravity's effect isn't increased over time, it is a constant force $\vec{f} = m \vec{g}$. $\endgroup$
    – Dimitri
    Mar 9, 2016 at 13:59
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    $\begingroup$ Doing some more googling, it does look like things drop faster at the end for bullets/artillery (en.wikipedia.org/wiki/External_ballistics) however, what you are saying is this is due to air/wind resistance or other factors? $\endgroup$
    – VenomFangs
    Mar 9, 2016 at 14:03
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    $\begingroup$ Okay, I see what you mean. The trajectory of any object (pointlike or of finite size) will always be parabolic in a constant force field if there is no friction. In the case of a football some shape effects coupled to air resistance might indeed modify the trajectory. $\endgroup$
    – Dimitri
    Mar 9, 2016 at 14:10
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    $\begingroup$ For something like a football, you can not neglect air resistance and that's why, indeed, the path is elongated on the throwing end and curved downward faster at the end. You also need to consider the Magnus effect, which allows football players to “bend” a shot in different ways. $\endgroup$ Mar 9, 2016 at 15:00
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    $\begingroup$ The faster drop at the end of a bullet's trajectory is caused by the bullet beginning to tumble instead of pointing directly forward. This is indeed caused by air resistance and loss of spin imparted by the barrel of the gun. A bullet fired in a friction-less environment would not have these issues. $\endgroup$ Mar 9, 2016 at 19:52

2 Answers 2


Many sources may tell you, that the path of a trajectory is a parabola. Indeed, all of the mathematical formula and calculations dealing with trajectories of objects falling, thrown or propelled support that interpretation. But when dealing with earth satellites and ballistic missiles, the truth is that their orbits are portions of ellipses.

The Newton's canon on a Mountain

Newton's cannonball was a thought experiment by Isaac Newton used to hypothesize that the force of gravity was universal, and it was the key force for planetary motion.

Newton's Cannon on a Mountain Here is an interactive version of it: Newton's Cannon on a Mountain

If an object has less than escape velocity (For earth it is 11.2 km/s), its path is an ellipse. If the object has velocity equal to escape velocity, it has a parabolic trajectory. If it is greater than escape velocity, it is hyperbolic.

Normally when we throw an object the actual path of the object is a part of a larger ellipse as the below image shows but since the velocity is not enough the object hits the ground before completing a full elliptical path which seems to be a parabola

enter image description here

The parabolic paths become flatter and flatter as the cannon is fired faster. Newton imagined that the mountain was so high that air resistance could be ignored, and the canon was sufficiently powerful.

PS: Newton's mountain was impossibly high but he realized that the moon's circular path around the earth could be caused by the same gravitational force that pulls cannonball in its orbit, in other words, the same force that causes objects to fall.

The answer to your question is that the path of football is truly "elliptical" since its velocity is way less than escape velocity. But to us, we "approximate its path to a parabola".

UPDATE: Mathematical answer to your question.

We can use equations of projectile motion as follows.

enter image description here

Equation for the trajectory of a projectile motion:

$\displaystyle y= x\tan\theta -{\frac{g}{2u^2\cos^2\theta}}x^2$

(yes it is an equation of parabola but I have mentioned earlier that the mathematical formula and calculations dealing with trajectories of object are approximated to parabola)

Now from your question we can have to situations:

CASE-1: When the object is thrown inclined at an angle $\theta_1$ with a velocity $u$

Then the maximum height the object will reach is given by:

$\displaystyle h=\frac{u^2\sin^2\theta_1}{2g}$

Now if $\theta_1= 30^\circ$ and initial velocity $u= 100\ \mathrm{m/s}$ (just for consideration)

Then the maximum Height the object will reach is equal to:

$h= 127.55$ meters

Now using the same angle and velocity, if we calculate the maximum distance traveled(called the range of projectile) we have

$\displaystyle R_\text{max}=\frac{u^2\sin2\theta_1}{g}$

Now by plugging in the values, we have $R=883.69$ meters

CASE-2: When the object is thrown at a higher angle than before but with same velocity.

Now say the angle $\theta_2=60^\circ$ (Higher angle than before) and $u=100\ \mathrm{m/s}$

Then by using the same equation used before we have

$h= 382.65$ meters and $R= 441.83$ meters


We can clearly see that the maximum height in case-1 is less than that of case-2 and the maximum range in case-1 is higher than that of case-2

Which means the path in case-1 is less high and more far. But the path in case-2 is higher and less far. See the below image for more clarification.

enter image description here (Sorry for the funky colors :P)

Image source: http://www.faculty.virginia.edu/rwoclass/astr1210/guide08.html; https://www.lhup.edu/~dsimanek/scenario/secrets.htm.

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    $\begingroup$ @VenomFangs "The velocity is more important than the angle considering the type of path(parabolic, elliptical or hyperbolic)". But since you asked about on earth phenomenon i would say that if the cannon was angled up slightly, the portion of the ellipse will look like "elongated at the start and curving down faster at the end" $\endgroup$
    – hxri
    Mar 9, 2016 at 14:49
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    $\begingroup$ This is a lovely answer. The only thing you could add (if you felt like it) is the calculation of the difference between the elliptical curve, and the parabolic approximation. This is something that would best be done numerically; I expect the difference is tiny (certainly much smaller than the difference due to ignoring effects of air drag and, in the case of a football, lift). $\endgroup$
    – Floris
    Mar 9, 2016 at 16:51
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    $\begingroup$ On the "parabola vs. ellipse" bit, if the Earth were flat, the football would, in fact, follow a parabolic trajectory. Since, on the scale of kicked footballs, the Earth looks flat, we can use the easier-to-calculate parabolic approximation rather than the exact elliptical solution. $\endgroup$
    – Mark
    Mar 9, 2016 at 21:36
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    $\begingroup$ 'The answer to your question is that the path of football is truly "elliptical" since its velocity is way less than escape velocity.' This is itself a weak gravity approximation. The real path of the object is to very high orders elliptical, but even neglecting air resistance etc., it is still not exact. $\endgroup$
    – Zorawar
    Mar 9, 2016 at 22:00
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    $\begingroup$ Absolutely right! Not many people aware of this fact. And we are all brainwashed in Physics 101 to believe the misconception of a parabolic flight path. Great presentation of the answer BTW. $\endgroup$
    – docscience
    Mar 10, 2016 at 15:35

To build up a little bit on our comments, do you mean that you expect the trajectory to be of different shape at the end than at the beginning ? This is again only possible with friction. If there is no friction, your Newton's equation of motion reads

$m \frac{d^2 \vec{r}}{dt^2} = m \vec{g}$

and is invariant under the transformation $t \mapsto -t$ (it is said to have time reversal symmetry). The practical consequence is that your trajectory must have some axis of symmetry : the football cannot move differently flying upwards than falling downwards because each of these movements map eachother through the time-reversal transformation.

However, if you add a friction term like

$m \frac{d^2 \vec{r}}{dt^2} = m \vec{g} - k \frac{d \vec{r}}{dt}$

(with $k >0$)

then you break time-reversal symmetry because $\vec{v}$ becomes $- \vec{v}$ under this transformation. Thus, as @leftaroundabout pointed out, you need a friction term to get the trajectory distorsion you described in your question.


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