This question already has an answer here:

Why does it take longer for an object thrown upwards to fall back down than it does for it to reach its highest point? We can draw a free body diagram for when the object is falling down and for when it's going up, but what does that tell us? Is there a simple argument for this or does one need to solve a differential equation? I first learned of this phenomenon during my first semester of physics but apparently I never really understood why it occurs.

enter image description here


marked as duplicate by sammy gerbil, M. Enns, Dvij Mankad, Jon Custer, Kyle Kanos Mar 12 at 14:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 5
    $\begingroup$ instead of "drag" in your diagram, think of buoyancy .Going up buoyancy is added to the initial impulse. In falling back, it works against the attractive force.. The extreme is a helium filled balloon, never falls down. $\endgroup$ – anna v Mar 11 at 6:46
  • 1
    $\begingroup$ Just wanted to make sure. I think Anna V is correct if the difference is very small or the density of the object is very low $\endgroup$ – Bob D Mar 11 at 7:20
  • 5
    $\begingroup$ Possible duplicate of Time of a ball going up and down with air resistance $\endgroup$ – sammy gerbil Mar 11 at 17:47
  • 1
    $\begingroup$ I suspect that the issue of buoyancy is leading people astray here. The combination of weight and buoyancy will define a natural acceleration for the particle. And we should use the direction of that acceleration as a reference rather than "up" or "down". We find that a particle thrown against the direction of natural acceleration has a lower speed on the return leg of a there-and-back then on the out-bound leg. If you throw it with the direction of natural acceleration it doesn't come back. $\endgroup$ – dmckee Mar 11 at 18:01
  • 1
    $\begingroup$ @Bob Buoyancy will effect the terminal velocity, but it will not affect the behavior in which the return leg is slower than the out-bound leg. lsusr correctly identifies the role that drag plays in the energetics of the situation (keep in mind that in the absence of cavitation we can reasonably expect buoyancy to be conservative and therefore treat it with a potential term making this analysis easy). $\endgroup$ – dmckee Mar 11 at 22:14

The Answer

Drag takes away energy from the object's total $E_g$ (gravitational potential energy) + $E_k$ (kinetic energy). Therefore the object's total $E=E_g+E_k$ is strictly decreasing.

Compare two points of equal height on the object's arc. The first point occurs while the object is rising. The second point occurs while the object is falling. Both points have the same $E_g$ because they're at the same height. The rising point has greater $E_k$ because $E=E_g+E_k$ is strictly decreasing with respect to time. Therefore for any pair of equal heights the object is moving faster when it's rising than when it's falling.

It takes the object an equal distance to rise as to fall. Since the object is moving faster (for any given height) when it's rising than when it's falling, that means the object takes less time to rise than to fall.

Note: Some comments to the original question mentioned the idea of buoyancy. Buoyancy does not cause an object thrown upwards to fall back down slower than it went up. That's because buoyancy is a conservative force, not a frictional force. Buoyancy force is constant with respect to time. All it does is decrease the effective gravitational acceleration.

A closer look at why buoyancy doesn't break symmetry

Several people seem to be confused about the effect of buoyancy on the thrown object, so let's take a closer look at the hypothetical situation where buoyancy acts on an object and drag does not. For simplicity, I will assume that the the ball is not thrown so high that the density of air appreciably changes. (If the density of air varies with $h$ then the math is harder but symmetry remains unbroken.)

There are two forces acting on the object. Gravitational force acts downward with a magnitude of $mg$ where $m$ equals the object's mass and $g$ equals the gravitational acceleration of an object near the Earth's surface. Buoyant force acts upward with a magnitude of $\rho Vg$ where $\rho$ is the density of air, $V$ is the volume of the object and $g$ is once again the gravitational acceleration of an object near Earth's surface.

The gravitational potential energy of a buoyant object in a medium of constant density is $E_g=mgh-\rho Vgh=(m-\rho V)gh$. At any point in the object's trajectory it's total energy is equal to its gravitational potential energy $E_g$ plus its kinetic energy $E_k$.

$$E=E_g+E_k=(m-\rho V)gh+\frac12mv^2$$

Solve for $v$.

$$v=\pm\sqrt h\sqrt{2\left(1-\frac{\rho V}m\right)g}$$

The variables $m$, $\rho$, $V$ and $g$ are all constant. That makes the expression $\sqrt{2\left(1-\frac{\rho V}m\right)g}$ constant too. In other words, the velocity $v$ is proportional to the square root of the height $h$ (up to a sign). That means the speed of the object is exactly the same when the object is ascending as when the object is descending for every pair of equal heights $h$. Without drag, the speeds of the object ascending and descending are the same. The displacement of the object ascending and descending is the same. Thus, the time $t$ it takes the object to ascend or descend is identical. They're both $t=\frac{2h}{v_0}$ where $v_0$ is the initial velocity of the thrown object. (This value comes from the fact that $\Delta x=\frac12(v_0+v)t$ for objects under constant acceleration. We know the object is under constant acceleration because the two forces gravitational force and buoyant force are both constant.)

It is true that the object will take longer to ascend and descend if you factor in a buoyant force. The question is asking what causes the ascention times and descention times to differ from each other. Buoyant force does not break the symmetry between a thrown object's ascention and descention.

  • 1
    $\begingroup$ I think both drag and buoyant force (which is always pointing upwards) are at play. $\endgroup$ – Alchimista Mar 11 at 9:06
  • 2
    $\begingroup$ @Alchimista As you point out, the direction of buoyant force is always upwards, and for a given object is a function of height only. It will have the same effect at the same height whether the object is ascending or descending. Buoyancy is a non-issue in this question. Direction of drag does change and affects the change in acceleration from ascending to descending more dramatically than anything else. $\endgroup$ – Bill N Mar 11 at 17:22
  • 1
    $\begingroup$ I am trying to reconcile this with the definition of "drag" which is opposite to the objects motion. simple.wikipedia.org/wiki/Drag_(physics)Buoyancy is opposite when going down but with the impulse going up. $\endgroup$ – anna v Mar 11 at 17:56
  • 1
    $\begingroup$ @Alchimista Accounting for buoyancy is no different from decreasing $g$. $\endgroup$ – PiKindOfGuy Mar 12 at 23:19
  • $\begingroup$ I have revised my answer in support of yours. For what it's worth, nice job! $\endgroup$ – Bob D Mar 13 at 1:18

@Isusr says the reason is air drag and that buoyancy does not affect the time. @Alchimista says the reason may be a combination of both.

Air drag does negative friction work on the object. Therefore it causes the object to loose kinetic energy cumulatively on the way up and down. So as @Isusr has said, it will have less kinetic energy (and therefore less velocity) at the same height on the way down than it did at the same height on the way up, yet the potential energy is the same at the same height. So the time up will be less than the time down due to air drag friction.


So what about buoyancy? Buoyancy force acts only upward, so I thought it should take longer to come down than go up with buoyancy than without buoyancy. But apparently I was wrong. I believe the jury has decided otherwise.

First, we have unanimous agreement that air drag causes the object to rise faster than it descends. That was concluded in the post that is cited as a duplicate. But that linked post did not address the potential influence of buoyancy.

Insofar as buoyancy is concerned, I am now convinced that it doesn’t cause the out bound trip to be in less time than the return trip. To me, at least, the compelling argument is that the buoyancy force only effectively reduces the weight of the object. I would prefer to think the buoyancy force effectively changes (reduces) the magnitude of acceleration due to gravity. I am looking a this from the perspective that both the buoyancy force and gravitational force are essentially constant for the distance travelled. All buoyancy accomplishes is to slow both the outbound trip and the return trip, but in equal amounts.

I see the OP has accepted @Isurs answer, and in my opinion rightly so. I have added my own upvote to the others. Kudos to @Isurs. And by the way, although the answer regarding drag may be a duplicate, the analysis by @Isurs of the possible contribution of buoyancy was not. So the answer, in my opinion, was not a duplicate.

Hope this helps.

  • $\begingroup$ Let me say that in this discussion I would omit significance, as for it depends on plenty of parameters. Density of the body and atmosphere. Shape. My only point is that while is correct that drag is a sink of kinetic energy up and down, buoyancy accelerates the body upwards. To the point that a body can even lift away, as @anna v said at the very beginning. I wasn't interested in quantifies things but to correct/discuss a wrong answer (in my understanding). $\endgroup$ – Alchimista Mar 11 at 20:17
  • $\begingroup$ @Alchimista Fair point. $\endgroup$ – Bob D Mar 11 at 20:54
  • $\begingroup$ I final comment for sake of clarity. @Isurs is right on the light of your answer. S/he is wrong on the way buoyancy is dismissed (a series of statement that do not hold) in both the answer as well as the subsequent comment. The best is to quantify the forces as you did, assuming we indeed speak of something much denser than the fluid through which is thrown. $\endgroup$ – Alchimista Mar 11 at 21:07
  • $\begingroup$ @Alchimista I'm afraid I need clarification on your clarification. You say Isurs is right on the light of my answer. Right about what? I thought he/she was right up until the note. Do you agree with the note? $\endgroup$ – Bob D Mar 11 at 22:04
  • $\begingroup$ Exactly S/he is right until the note. Though a speed/drag non linear dependence can even complicate the calculation, the fact that drag is an energy sink seems to point to a slower fall down as compared to climb . And something thrown very high might stay longer in thin atmosphere while falling. ..but is a bit out of the argument. $\endgroup$ – Alchimista Mar 12 at 8:15

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