I noticed that if you drop a flat surface, parallelly from a relatively low height, onto another, the impact is almost cushioned.

What is happening here?

Is this to do with the air, especially near the centre, not being able to escape, thus applying a repulsive pressure?

  • $\begingroup$ In fluid mechanics it is known as 'drainage'; the emptying of air between two interfaces before they touch. It is for example an important phenomenon in coalescence of droplets in air where they can only coalesce when the air between the droplets has drained. Often this is the determining step of the process. $\endgroup$
    – nluigi
    Commented Mar 4, 2017 at 9:20

2 Answers 2


What is happening is that there is a volume of air under the thing being dropped, which has to make its way out the sides. As the distance gets smaller, the air pressure under the object increases while the air is escaping.

Try a flat board, then drill some holes in the board that account for maybe 10% of its area. Despite being 10% lighter, the board will then land much faster since you have provided a lot more area for the air to escape.

Of course in a vacuum, no difference and even a very light flat object would just slam down.

  • $\begingroup$ How does your explaination differ from mine? $\endgroup$
    – Tobi
    Commented Mar 3, 2017 at 20:13
  • 4
    $\begingroup$ @Tobi I think he's confirming your hypothesis -- that's how it works, so it doesn't need a "different" explanation. Your explanation is largely correct, so he's simply restating and expounding upon it a bit. macgyver_sc provides the sciencey explanation in his answer, but this is a perfectly accurate explanation unto itself. $\endgroup$
    – Doktor J
    Commented Mar 3, 2017 at 20:45
  • 1
    $\begingroup$ @Dira: The object is falling faster than the air can escape. The air has to go somewhere. Therefore, it is compressed. $\endgroup$
    – user131054
    Commented Mar 3, 2017 at 22:37
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    $\begingroup$ @Diracology The basic Bernoulli equations only apply to imcompressible fluids in steady flow where effects of viscosity are low. I think it's application here is very limited. $\endgroup$
    – JMac
    Commented Mar 4, 2017 at 2:47
  • 1
    $\begingroup$ @DoktorJ His (unedited) answer started with the statements "I disagree. It's similar, but there's no aerodynamic lift involved." which - confusingly - implied contradiction. $\endgroup$
    – Tobi
    Commented Mar 4, 2017 at 19:14

That can be explained using Fluid Mechanics and Hydraulics, because air is a fluid. Torricelli's Law gives us the expression of the total time a fluid needs to empty a container. It is:

$$ \Delta t = \frac{2A}{a\sqrt{2g}}(\sqrt{h_1}-\sqrt{h_2}) $$

In your case, there is no container, but the paper's weight has a force by area over the fluid air, what creates pressure. The air as a fluid needs time to displace and depressurize, what creates a force that holds the paper. If you did the same in vacuum you wouldn't have the same results.


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