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I was once asked the following question by a student I was tutoring; and I was stumped by it:

When one throws a stone why does it take the same amount of time for a stone to rise to its peak and then down to the ground?

  • One could say that this is an experimental observation; after one could envisage, hypothetically, where this is not the case.

  • One could say that the curve that the stone describes is a parabola; and the two halves are symmetric around the perpendicular line through its apex. But surely the description of the motion of a projectile as a parabola was the outcome of observation; and even if it moves along a parabola, it may (putting observation aside) move along it with its descent speed different from its ascent; or varying; and this, in part, leads to the observation or is justified by the Newtons description of time - it flows equably everywhere.

  • It's because of the nature of the force. It's independent of the motion of the stone.

I prefer the last explanation - but is it true? And is this the best explanation?

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    $\begingroup$ All three things you say are correct, possible answers. Asking why? is not really a good question, and it is unclear what kind of answer you want to hear. What would a "best explanation" be? $\endgroup$ – ACuriousMind Feb 15 '15 at 14:07
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    $\begingroup$ @acuriousmind: I think this depends on what one considers as a good explanation; I've indicated what I think is the best explanation - but I'm not even sure that is correct. $\endgroup$ – Mozibur Ullah Feb 15 '15 at 14:23
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    $\begingroup$ I thought asking why was partly the point of physics. $\endgroup$ – Mozibur Ullah Feb 15 '15 at 14:24
  • $\begingroup$ Why does it take a projectile as long to get to its apex as it does to hit the ground? Only true if the projectile is launched from the ground. With an upward trajectory. At least for me, knowing the constraints under which the condition hold true explains the why. $\endgroup$ – HopelessN00b Feb 16 '15 at 2:29
  • $\begingroup$ This is trivially true in the absence of air resistance, but (not so trivial) untrue if air resistance is present. Air resistance is not time-symmetric, of course, which explains how @By Symmetry's answer is right given the unstated assumption. $\endgroup$ – MSalters Feb 16 '15 at 13:43

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I think its because both halves of a projectile's trajectory are symmetric in every aspect. The projectile going from its apex position to the ground is just the time reversed version of the projectile going from its initial position to the apex position.

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  • $\begingroup$ I like this answer because it ties in with symmetry - which is a fundamental notion now. $\endgroup$ – Mozibur Ullah Feb 15 '15 at 15:49
  • $\begingroup$ This seems like an identical answer to the one posted by @BySymmetry. One of you must not have noticed the others' answer... $\endgroup$ – Floris Feb 15 '15 at 19:26
  • $\begingroup$ I thik this is the best explanation; Newtons laws are time-reversible, and gravity is a constant force for projectile motion. I should have accepted this some time ago ;). $\endgroup$ – Mozibur Ullah Aug 24 '16 at 19:04
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I would say that it is a result of time reversal symmetry. If you consider the projectile at the apex of its trajectory then all that changes under time reversal is the direction of the horizontal component of motion. This means that the trajectory of the particle to get to that point and its trajectory after that point should be identical apart from a mirror inversion.

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  • $\begingroup$ This seems like an identical answer to the one posted by @HritikNarayan. One of you must not have noticed the others' answer... $\endgroup$ – Floris Feb 15 '15 at 19:27
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    $\begingroup$ @Floris: that's pretty common, isn't it? For simple questions like this, at any rate, that can be answered quickly. The answers will just have been posted at about the same time. (It's happened to me more than once over at Stack Overflow.) $\endgroup$ – Harry Johnston Feb 15 '15 at 22:51
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I would consider that since acceleration is a constant vector pointing downward, that the time the projectiles downward component takes to accelerate from V(initial) to 0 would be the same as the time it takes to accelerate the object from 0 to V(final)

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  • $\begingroup$ This is probably the easiest way to explain it to someone in an intro to physics class, as long as you are able to convince the student that v(final) = -1 * v(initial). $\endgroup$ – mbeckish Feb 16 '15 at 14:50
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A lot of things have to hold to get that symmetry. You have to neglect air resistance. You either have to throw it straight up, or the ground over there has to be at the the same altitude as the ground over here. You have to through it slowly enough that it comes back down (watch out for escape velocity)

But if you have that, then the simplest explanation in the simplest case is that the acceleration is constant. So you keep going up until you have zero (vertical) velocity, which takes time $v_0/g$, during which you travel up a certain distance. On the way down, the acceleration is still constant, so now the velocity becomes more down over time, but that just makes it the time reverse (when looking at the vertical component alone) so it takes an equal amount of time.

If you want to throw it high enough that the acceleration starts to vary, an energy argument is quicker, though you can study the kinematics on the way back down as the time reverse (gaining just as much as you lost before, in just as much time, going just as far, etc.).

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  • $\begingroup$ While this answer does require an understanding of acceleration, I feel it's the best because it doesn't require any understanding other than the basics. None of this "time symmetry" nonsense that fails as soon as you add friction: just pure "plot on a graph where a ball at 0,0, traveling up at 100m/s and decelerating at 10m/s will be after each second, until it passes the graph axis." $\endgroup$ – Dewi Morgan Feb 16 '15 at 2:12
  • $\begingroup$ After a while, you understand that you're adding the exact same amount to the vertical distance each step on the down-stroke, that you removed at the same y-value step on the up-stroke. So even varying g becomes easy. Then you add in friction, and terminal velocity, and suddenly the graphs look asymmetrical and the students understand why. $\endgroup$ – Dewi Morgan Feb 16 '15 at 2:16
  • $\begingroup$ Actually, only the air resistance breaks the symmetry. Different heights for landing and takeoff don't matter - they just alter the initial and final velocities. (Imagine lobbing a ball up so that it just lands on a ledge. That's the time-reversed version of nudging a ball off the ledge and it falling to the ground). As for escape velocity, an object that flies off to the edge of the universe is a time-reversed object that falls in from the edge of the universe and hits the ground at escape velocity. $\endgroup$ – Oscar Bravo Feb 16 '15 at 13:21
  • $\begingroup$ You don't have to throw it straight up. $\endgroup$ – Solomon Slow Mar 7 '15 at 17:42
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One could say that this is an experimental observation; after one could envisage, hypothetically, where this is not the case.

This is not hypothetical once you take air resistance into account.

One could say that the curve that the stone describes is a parabola; and the two halves are symmetric around the perpendicular line through its apex. But surely the description of the motion of a projectile as a parabola was the outcome of observation; and even if it moves along a parabola, it may (putting observation aside) move along it with its descent speed different from its ascent.

Assuming constant horizontal velocity, the height vs time diagram will essentially look the same as the height vs horizontal distance diagram. Under that assumption, it is valid to argue about time-of-flight from the symmetry of the trajectory; in general, it's not.

It's because of the nature of the force. It's independent of the motion of the stone.

The motion of the stone depends on the force, so it's hardly independent.

Under constant vertical acceleration, the change in vertical velocity is linear in time. This means that it takes the same amount of time to go from initial velocity $+v$ to 0 as it takes going from 0 to $-v$. In fact, the velocity profiles will be mirror images, and the same distance will have been covered.

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  • $\begingroup$ note to self: verify that ascent and descent actually do not take the same amount of time once you add drag... $\endgroup$ – Christoph Feb 15 '15 at 15:18
  • $\begingroup$ Both drag and gravitation act to slow down an ascending object. Drag still acts to slow down a descending object, but gravitation acts to speed up a descending object. Speed decreases faster during ascent than it increases during descent. Another way to look at it: With drag, the speed at some time $t$ before apex will be greater than the speed at that same time after apex. Yet another way of looking at it: Ascent takes less time than descent once you add drag. $\endgroup$ – David Hammen Feb 15 '15 at 17:50
  • $\begingroup$ @DavidHammen: I'm slightly ashamed to admit it, but I actually did have to draw some pictures to convince myself that this is valid reasoning ;) $\endgroup$ – Christoph Feb 16 '15 at 13:46
  • $\begingroup$ I'm even more ashamed to admit it, but I actually had to do the math to ensure I was right before I wrote my previous comment. Sometimes there's a lot to be said for a qualitative answer. $\endgroup$ – David Hammen Feb 16 '15 at 13:56
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(Non-kinematics math attempt but just some principles)

It is a partial observation in that

  • It hits the ground with same speed.
  • Angle by which it hits the ground is the same (maybe a direction change)
  • It takes equal time to reach to the peak and then hit the ground

They are equally strange coincidences. Which are more fundamental? Consider the following setup: There are two smooth surfaces of a mountain with different angles. You kick a ball from the ground along a surface towards the hill top so that it just reaches the peak. Now if it rolls back from the same surface, all the above entities will be preserved (including the angle as the ball rolls down on the same surface). But if the ball just goes over the peak (with almost zero speed) and rolls down via the other surface, it will reach the ground at in shorter or longer time and with a different angle. But the speed will be the same (conservation of energy!).

So, once an entity is conserved, there we can find a symmetry. The opposite is also true. If there are symmetries, then some thing must be preserved. I heard this is proved mathematically long ago.

In your setup you have all the symmetries and it is no wonder you have conserved entities in the process.

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    $\begingroup$ If there are symmetries, then some thing must be preserved. I heard this is proved mathematically long ago. — Yes. You are thinking of Noether's theorem. (The word usually used is conserved, not preserved.) $\endgroup$ – zwol Feb 15 '15 at 22:31
  • $\begingroup$ Thanks for noting...It is wonderful how math and physics fit together in nice ways... $\endgroup$ – mami Feb 15 '15 at 22:49
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When one throws a stone....

Your arm is capable of propelling an object at up to around 150km/h (and that's with some practice). At that speed the many factors like air resistance are negligible.

Let’s load a 16-inch shell into a gun (you will find several on the USS Iowa), aim it 45 degrees up and press the button. The shell is going to go up at an initial rate of Mach 1.7 (vertical vector), decreasing to zero at apogee, and it will come down accelerating at g until it reaches its terminal velocity. Things tend not to fall at supersonic speeds so it should be clear that the up time will not equal the down time.

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Asking "Why" in physics often leads out of physics and into philosophy. Why is there light? Because God said so. Physics only answers questions about how the universe behaves and how to describe that behavior. If a why question can be answered with physics, the answer is "because it follows from a law of physics." A law is just a mathematical description of behavior that has been verified with experiments.

In this case, the answer follows from $F = ma$. If $a$ is constant, this leads to $x = 1/2at^2 + v_0t + x_0$. Since you are a physics tutor, you can set up the initial and final conditions and solve for t.

This may well lead to more questions. Why is $F = ma$? Why is there gravity? You could fall back on philosophy, or you could say "I don't know. But experiments show that it works this way."

Why questions are usually about things that are unfamiliar. A young child will ask "Why is there gravity?" A high school student is more likely to ask "Why is $F = ma$?" He may not know the answer to the first question, but he has become used to it.

So the way to answer $F = ma$ questions is to show how it applies to familiar situations. This is why high school physics is full of pictures of springs and ropes and pulleys.

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    $\begingroup$ I'd suggest that part of the training of physics is to understand the appropriate kind of why; the why that is part of its domain of investigation; and its explanatory principles. $\endgroup$ – Mozibur Ullah Feb 15 '15 at 15:48
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Assuming that the only forces acting on the stone are the initial force exerted by the hand on the stone AND the force of gravity, we have the following situation:

Immediately after throwing, the stone has a kinetic energy in the x-direction that never changes (conservation of energy) and the stone has a kinetic energy in the y-direction that is constantly being depleted by gravity as it rises to its apex. Once the stone is at its apex (vertical kinetic energy = 0), gravity works to restore all of the stone's original vertical kinetic energy as it falls back to its original height.

So, the x-direction kinetic energy remained constant throughout the flight and the y-direction total energy (kinetic + potential) remained constant throughout the flight, independent of each other. This combination causes the symmetry of the situation.

If air resistance and terminal velocity are introduced into the problem, the character of the problem is completely changed. Even a very simple situation as the one you described can become very complex and confusing to most people. Most people would be stymied by the complexity of the math involved even in this simple situation, let alone the complexity of putting a probe into orbit about Earth.

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Conservation of momentum!

Force × Time = Impulse = Δ Momentum

Since the average force is the same going up and down, and since the momentum change is the same going up and down as well, the time during which the force is applied must also be the same.

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protected by Qmechanic Feb 15 '15 at 22:59

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