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The scene where the A-Team is in the tank and its falling from the air, they fire the canon and it stops from falling for a moment before falling again. Is this possible from a Physics point of view?

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    While this seems to be absolute crap (regarding physical reality, of course, not regarding the A-Team, in the movie this scene was just amazing), I heard that someone checked if this is possible and came to the conclusion that it might work. But I have absolutely no sources on this currently. That being said, dropping a tank on parachutes is not that uncommon or impossible in the first place. – Napoleon Wilson Sep 10 '14 at 23:23
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    This looks like a good question for Randall Munroe's What If?. – Vedran Šego Sep 10 '14 at 23:52
  • This is really a question for physics.se – DA. Jan 28 '16 at 17:46
  • This has already been posted to physics se and answered. But I didn't realise it was still here. Should I take it down? – Aasim Azam Jan 29 '16 at 15:41
  • this has already been answered physics.stackexchange.com/questions/134943/… – Aasim Azam Apr 29 '16 at 22:23
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Let's do some sums. I can't find specs for the main gun, but a similar British Army 105mm light gun fires shells weighing 15.1kg at 709m/s. Using 1/2mv^2, that's 3.80MJ. Let's assume all of the energy from the firing went into the projectile when the tank was on solid ground, because that's the worst case - anything else would improve our chances of stopping the tank. Since the tank and projectile are now free-floating, half the force (and hence half the kinetic energy) goes to the tank.

The lightest M8 tank was 19.25 tons, which is 17,463kg. If the tank was falling at 14.7m/s (53km/h), one shell would stop it completely. From there on, doubling the speed needs four times the shells - so 29.5m/s needs four shells, 59m/s needs eight shells, and so on.

That's discounting acceleration from gravity between shots, of course. The main gun manages 12 rounds a minute, which is 5s between shots. If the tank is free-falling, that's 9.8m/s/s acceleration, which clocks up 49m/s between shots. So even if they weren't moving when they started firing, they couldn't keep the tank up. It seems unlikely that the single remaining chute (after the Reapers had shot the rest) would slow the tank enough that the gun could even keep up with gravity, never mind bring the speed down enough for a survivable impact on the lake.

  • If you found a typo in your answer, you can as well just fix exactly that instead of adding a line that says there is a typo. I would correct it, but I have no idea which of all those numbers is supposed to be 49m/s. – Napoleon Wilson Feb 1 '16 at 11:27
  • @Napoleon Wilson Fair point. I just have a dislike of articles that change underneath you! Now fixed, and the edit note says what I changed. – Graham Feb 1 '16 at 13:36
  • I don't see how this is in any way a signifcant change at all, though. And besides that, there's always the revision log to track changes. – Napoleon Wilson Feb 1 '16 at 13:40
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The physics at work here are compression waves and terminal velocity.

Think of a large speaker cabinet. If you stand up a piece of paper a few feet away and put the volume of your stereo all the way up, the paper will be blown over. Much like Sonar, the compression waves move air toward an object.

Now, to stop a steel tank, those waves would have to be incredibly powerful. It's not likely that compression waves from a tank cannon could generate enough force to stop or even slow down an object that likely weighs a couple tons travelling at terminal velocity.

  • What about recoil? – Vedran Šego Sep 11 '14 at 3:47
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    @VedranŠego No. You'll slow down the tank a tiny bit, but only by a marginal amount. You obviously heard about actio = reactio, so E_projectile = E_recoil applies, but don't forget the massive mass difference between the tank and the shell. If you see a tank shooting and there's visible recoil (like the tank shaking), then this is really just due to the fact that this recoil doesn't have to directly work against the tank's own weight. – Mario Sep 11 '14 at 6:41
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    No, the only relevant physics is recoil. @Mario is wrong about the mass difference. Recoil has two factors: one is mass, and the other is the acceleration of the projectile (F=ma), and that force is directly applied to the tank which experiences F=ma in the opposite direction. Terminal velocity would be relevant if the tank reached it, but they're trying to slow it down so it isn't. – Graham Jan 28 '16 at 12:18
  • I'm not a physicist, but I'm pretty sure this is incorrect. As Graham states, it's recoil at work. The physics here: bsharp.org/physics/recoil – DA. Jan 28 '16 at 17:45
  • Also, I don't believe the tank ever hits terminal velocity as it has parachutes...it just reaches a speed faster than the human body would appreciate upon landing. – DA. Jan 28 '16 at 17:45
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Tank guns have the technology to compensate the Recoil of the fired shoot.

This Wiki article is about recoil and mordern ways to compensate it.

And this article is about the Muzzle brake - also to minimize the recoil.

Both technics are straight against the fact that a fallen tank can be slowed down by shooting its gun. It would only start to spin around.

The only other thing I could think of that could (very unlikely) help them is the therorem of self lifting bodies.

But not only therefor the Tank must have gotten rid of the parachute, it also needed to get a bit more round on its edges.

  • Recoil and compression waves are not the same. When the projectile is fired, it still pushes air towards the object. – Johnny Bones Sep 11 '14 at 16:37
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    This xkcd "what if" discusses the lift created by various machine guns and might be applicable reading material. what-if.xkcd.com/21 – Leatherwing Sep 11 '14 at 20:34
  • The recoil has to go somewhere, though (I believe). On the ground, that force would be absorbed by friction...which there is much less of in the air. – DA. Jan 28 '16 at 17:47

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