# How much force would a pistol require to keep someone airborne? [closed]

I want to have a villain packing a pistol that he can fire upwards at the 'hero'- who blocks it with the flat of his sword to prevent himself from falling downwards.

The pistol is to move the hero briefly upwards before he then returns to his previous height when he blocked the bullet. It should only take perhaps half a second to fall back to the previous height.

Assume the 'hero' is average weight and has a 3 kilogram sword.

Assume the bullet from said pistol travels at supersonic speeds or above.

Assume the hero has been falling for 1 second.

How powerful would this pistol have to be? How much force would it have to produce? What would be its effects on the human body for the person firing and the person hit if it were to hit?

• This depends on how fast the person is falling. If they're at terminal velocity, the bullet would have to have a lot more momentum than if they had just jumped off a ledge or something. Commented May 7, 2017 at 19:05
• Fair enough, forgot about that. Commented May 7, 2017 at 19:12
• The a single bullet cannot (under normal physics) "just prevent him from falling for a second or so". It could stop him, though he would begin falling instantly. It could push him up enough that it takes the hero a second to fall back to his previous altitude. Gotta be one or the other. Commented May 7, 2017 at 19:12
• @Xavon_Wrentaile I added the stipulation that it pushes him up enough that it takes half a second to return to his height before blocking the bullet. Commented May 7, 2017 at 19:15
• Don't be so modest: what-if.xkcd.com/18 . Or directly give the gun to your hero: what-if.xkcd.com/21
– Karl
Commented May 7, 2017 at 20:08

It does not work.

Newton third law already says it: If the pistol can levitate something, it is exactly the same amount of force the wielder would carry. And no, even Andre the Giant cannot hold a falling person with one hand.

The second problem is the difference of energy and momentum. To move something, you need momentum, the product of mass and velocity.

The most powerful .357 Magnum bullet weighs 12 gram (let's say the villain is serious). A falling body of 70 kg for half a second means it fell 1,25 m (0.5*g*t^2) and achieved a velocity of 5 m/s.

This means the body has now a momentum of 350 Ns. A bullet with the same momentum must have a velocity of 350 / 0.012 = 29 166 m/s which is as fast as a meteor.

Even if you accelerate a bullet to such speeds, it has a much, much smaller surface area than a sword or a shield. This means the velocity is not used to accelerate the body backwards, but to penetrate the sword/shield. There is also no option to prevent this because the bullet is faster than the internal sonic speed, it moves inside the sword molecules before they have time to yield.

The thing is that "small fast thing moves big slow thing" also occurs in comedies and comics because it looks funny. The reason is that it is funny because it does not occur. If the big mass is n times more heavy than the small mass, for every gain or loss of 1 m/s of the big mass the small mass must move n m/s slower.

Big mass protects. If I put myself on the ground, holding an anvil over my belly, you can send the most vicious badass with a sledgehammer to pound the anvil and exactly nothing will happen. If a truck and a car collides head on, the truck is moderately slowed and feels a small impact, the car in contrast will be completely destroyed and driven back.

• 'And no, even Andre the Giant cannot hold a falling person with one hand.' Technically everyone is 'falling' towards the earth, and he definitely could hold someone with one hand. Further, parents regularly catch their falling child after throwing them in the air. I think this needs to be clarified with how long the person had been falling before he tried to grab them.
– Rob
Commented May 7, 2017 at 23:40

Let us make it super easy, and do the math. Because if I am going to do the math it needs to be super easy.

The hero, robust like myself, weighs 120kg. He is going to fall under earths gravity 9.8 m/s2. I say going to fall because I want him to have velocity 0, not the velocity after falling for 1 second because I don't want to have to add his momentum to the continuing acceleration of gravity.

The bullet will stop him from falling. It will do so by imparting a vector force in the opposite direction of gravity's acceleration and exactly equal to the acceleration of earth's gravity on his mass. We will assume a perfectly elastic collision. We will ignore the acceleration of gravity on the bullet. How fast must the bullet go to impart that force?

F=ma which is easy. For the hero and gravity m is 120 kg and 9.8 m/2 and so F is 1184.4 newtons acting on the hero as a result of gravity.

A shotgun slug is 28 grams and the Desert Eagle fires bullets that are 21 grams. Let us say the bullet weighs 3 grams which is a bigass bullet. 3 grams = 0.03 kg F = 1/2mv^2. Solving for v

• 1184.4 = 0.5 * 0.03 * v^2

• 1184.4 = 0.015 * v^2

• 78960 = v^2
• 280 = v

280m/s is not fast for a bullet. I found velocities of 300-400m/s for shotgun slugs on the web. I am surprised and I feel I must have left something out or misplaced a decimal.

So it should be possible to fire a bullet which, in an elastic collision, counters the force of gravity on a falling human. A faster bullet or bigger bullet at the same speed should more than counter the force of gravity.

My math is weak. But this math seems right and if it is wrong I want to be corrected. @supercat I am sorry to impose but would you please look at this math and correct it in comments? You used to do some math on the halfbakery.

• 3 grams ≠ 0.03 kg Commented May 7, 2017 at 22:27
• I think the math is correct (except for the mass as Daniel Beck pointed out) but it's just the velocity required to keep a "just starting to fall" person at the current position. That doesn't answer the question - the bullet not only needs to stop him falling (for a single instant) but to accelerate him upwards. Then the velocity is a bit different. Commented May 8, 2017 at 0:05
• Thank you @Daniel Beck. Those pesky orders of magnitude. So the answer is 2800 m/s. The fastest rifle bullet I found was 1112 m/2. Commented May 8, 2017 at 0:53
• @MSeifert I wanted to do the simpler thing and just see if it was possible for a reasonable bullet to oppose gravity on him. You are right: the OP wants more. He asks for him to be moving downwards and turn around and move upwards which would mean a faster bullet. Commented May 8, 2017 at 0:53
• @Will it was originally just to stop the hero in midair but one of the comments made it seem like that wasn't feasible so I changed the question. Commented May 8, 2017 at 4:45

A lot of force would be required, the villains gun would probably be more of a cannon than a gun. The most likely event is the sword would be torn out of the hero's hand and his wrist broken.

If the supposed "bad guy" in your theoretical situation was using something along the lines of recoiless rifle/pistol then he would not experience the massive amount of force that he was putting on target. Very similar to a rpg7 where there is a back-blast. Mind you it would have to be a very strange weapon but I figure that would get around your hand held issue.

Another alternative would be to have reduced gravity during this event. Think John Carter if you will. I hope that this is helpful in some way.

If instead of the pistol the shooter has something like an energy cannon which will project continuous force towards the target while the trigger stays triggered, then you could conceive of a shield which can spread this energy around its surface and dissipate the energy enough for it to become a manageable force and push the target away from the shooter (he is essentially snowboarding at this point).

This scenario allows for the speed of your target to be decelerated, allows for a short reaction time so he can put the shield between him and the beam, and the energy to stay long enough so he can be accelerated in the reverse direction.