We need our OP MaxTheVool to “weigh” in with his ruling on this. If you throw an object downward with some force (in addition to gravity) it will bounce higher than if you don’t. Does that count?
If I’m not mistaken, an object that falls solely under the force of gravity can never bounce higher than the originating altitude, and that only if there is zero air resistance and perfect elasticity.
I read a children’s sci-fi-ish novel once called Peter Graves (1950) (apparently unrelated to the actor of the same name), about a stereotyped “crazy professor” who invented a substance that would bounce higher than the originating height. (This was before flubber, BTW.) Peter Graves was a teen-age kid who stumbled upon his secret, with adventures ensuing.
Their equation includes mass because mass is relevant to terminal velocity: mass determines how much force is being exerted by gravity. When the object has accelerated to high enough speed where the aero drag exactly offsets the gravity force, the object stops accelerating; it has reached its terminal velocity. That’s why mass has to be part of that equation.
If aero drag is neglected, then the only force remaining is gravitational attraction, and believe it or not, the object’s mass falls out of the equation. The gravitational force on an object is proportional to its mass:
F = m * a
where:
F = gravitational force pulling an object down toward the earth
m = mass of object
a = acceleration due to gravity (9.81 m/s[sup]2[/sup])
But the acceleration of an object is determined by the gravitational force acting on it, and by its own mass, thus:
a = m/F
It should be clear at this point that the downward acceleration of the object a is a constant, regardless of the object’s mass or velocity (again, no force involved except gravity in this scenario). It’s always 9.81 m/s[sup]2[/sup], unless you start to get meaningfully far from the earth’s surface.
So for an object released at zero relative velocity at time t=0 and acted upon by gravity alone (no aero drag), the velocity as a function of time is this:
V(t) = a * t
where:
a = 9.81 m/s[sup]2[/sup]
t = time, s
V(t) = downward velocity, m/s
Air resistance acts during the entirety of a fall, not just once you reach terminal velocity. In fact, it takes a theoretically infinite amount of time to reach terminal velocity: You just approach it asymptotically. So if you have a solid cannonball and a hollow one of the same size, and drop them both at the same time, the solid one will immediately take the lead, not only after dropping for a bit. It’s just that the lead is going to be extremely small at first.
I suppose someone might have pointed out the problem of bodies falling at the same rate early on.
The other problem is that at terminal velocity steel despite it’s ductile qualities is going to permanently deform when it hits another piece of steel at terminal velocity and lose a lot of it’s ‘bounce’. To get the highest bounce you need the material with the most resilent nature, something that will deform and resume it’s initial shape again.
The OP didn’t specify shape or complexity of the object, so we could abandon our obsession with solid spheres and go with a finned, aerodynamic mass with a huge spring or gas cylinder on its nose. If the vehicle’s mass is large compared to the mass of the spring (or of the piston in the gas cylinder), then you could have an extremely high-energy, extremely elastic collision; the rebound could be very high.
You would need a specially prepared landing pad, something that could handle the impact force with minimal deformation or displacement, so that it would not absorb much of the energy at all. A steel bedplate with a mass many times that of the projectile, set in concrete, ought to do nicely.
Alternatively you could have the spring/gas-cylinder built into the landing pad, but that would require precision aim; the former configuration allows you to simply build a large landing pad and have the projectile land somewhere on it.
Yes - I know what terminal velocity means and why objects falling through a medium do not accelerate without bound.
I suspect all of us posting serious replies in this thread have a sufficient understanding of the trivial case in which an object falls through a vacuum or drag is neglected for the sake of simplicity as a first approximation.
And while the effects of drag may be small and might safely be neglected in some circumstances, the effect is obviously not zero and that is the point that I was addressing.
I was responding to a post in which a claim was made that seemed to be at odds with the actual physics involved when drag is considered.
Thanks - this is what I was getting at in my post.
It seems odd to me that someone would acknowledge that mass affects the terminal velocity of an object falling through a medium but would not affect the rate at which it gets to that speed.
I was starting to think in that direction. I wondered how high the craft that made balloon landings on Mars and Luna bounced. A ball made of very tough material with a compressed gas inside might be the simpler answer without requiring complex mechanics, and it could land on any hard surface because it could be very large and the force of the impact spread over a large area. A tear drop shape could allow it to attain very high speed and directional stability in an atmosphere, and dirigible* type fins could be used also.
I think the above two posts misunderstand the term “terminal velocity”.
Terminal velocity is where gravitational force matches wind resistance, thus reaching maximum free-fall velocity. As Chronos notes above, if you’re just falling, you never quite reach it, and if you were reentering at greater than terminal velocity, you’ll slow down until you (nearly) reach it (if you fall long enough). So, the initial velocity matters only if it’s really high compared with the terminal velocity; otherwise the two cases are nearly the same speed (the asymptotic terminal velocity). Furthermore, there are plenty of cases (e.g., feather) where the initial velocity is nearly irrelevant; it’ll reach terminal velocity very quickly, unless it’s moving fast enough to burn up, in which case the discussion is irrelevant.
I also bet it’s the case (as Chronos says) that for a solid metal ball the size of a cannonball dropped from airplane heights, terminal velocity isn’t reached.
Well, yes, because terminal velocity (assuming ideal conditions) is never reached. What people usually mean when they speak of “how long to reach terminal velocity” is “how long to get close enough to terminal velocity that it’s almost there”. The problem with this question, though, is then you have to define how close is “close enough”. 90% of terminal velocity? 99%? Those will give very different answers.
OK, the idea of using compressed gas as our bouncy medium got me to thinking. We’d want something like a compression cylinder, filled with an ideal gas. Compress the cylinder adiabatically until it’s to the point where it doesn’t behave like an ideal gas any more, and let it rebound from there. Take the maximum amount of energy you can store that way, and convert it to gravitational potential energy to find a bounce height. We clearly don’t want any parasitic mass, here: We’ll get the greatest height if the only mass involved is that in our energy-storage medium. So we can just reduce it to the single-molecule level: How much energy can you store per molecule of an ideal gas before it ceases to be idea, and how high will that get you given the mass of that molecule?
I’m pretty sure that we’ll get closest-to-ideal behavior from helium. That won’t cease to be ideal until the energy per atom gets to the ionization energy, or 24.59 eV. Using the simple formula for gravitational potential energy, E = mgh, that gets us to a height of 60,000 km. Which is obviously well outside the range of applicability of the simple energy formula. I don’t feel like going through the more detailed calculation right now, but I’m pretty sure that we’re well over escape energy, here. In other words, with a sufficiently efficient bounce system, and neglecting atmospheric drag, it looks like we could make something bounce as high as we’d like.
Ah - no I was talking about the net acceleration of the falling object when drag is considered.
Also, the equation that I referenced was actually velocity as a function of time v(t) which includes terminal velocity v[sub]t[/sub] as the first term with the second term being the tanh factor that calculates what fraction of terminal velocity has been reached at some time t.
And I’m not trying to pick a nit here just for the sake of it, but the notation you’ve used above:
seems to indicate terminal velocity as a function of time, which of course it is not.
I know that you know that, but it is what you wrote. And things have gotten a bit muddled in this discussion so I wanted to point it out in case it confused someone else following along.
Apologies if I wasn’t clear enough in my previous posts and it led to some misunderstanding about my meaning.
That can’t be right. Are you saying that if someone jumps out of a plane very high up, then opens up their parachute, then parachutes earthward for 10 minutes, their velocity at minute 3 of parachuting is actually different than their velocity at minute 5 or minute 7 or minute 9?
I feel like that must be one of those things that would be true in some hypothetical “ideal gas”, but not true in an actual atmosphere.
According to the simplest mathematical model for velocity as a function of time when falling through a liquid medium of constant density, it’s true: your velocity asymptotically approach terminal velocity.
The atmosphere is not constant density, though. The reality is that if you’re descending under canopy for ten minutes, then in the tenth minute you’re probably descending more slowly than in the first minute, because the air is more dense (and so causes more drag) at lower altitudes. One very extreme example of this was Felix Baumgartner’s jump last year, where he lept from the gondola of a balloon at 120,000+ feet. He exceeded the speed of sound early on, and then - as he descended to the more dense layers of the atmosphere - he decellerated to just a couple hundred MPH, at which point he finally deployed his parachute.
Chronos was talking about the theoretical model, and assuming a spherical body.
In reality, not only does terminal velocity change as you descend, it can change radically as your body moves or reorients as you fall (if you’re not a sphere or are not aerodynamically stable).
In cases where terminal velocity is approached quickly, and where it isn’t a constant … well, there’s enough slop there to say “it reaches terminal velocity” under a reasonable definition, and also say “it doesn’t quite reach terminal velocity” under another reasonable definition. (The former being an average TV, the latter being the instantaneous TV.)