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So I know how to do the math of instantaneous speed when doing free fall:

$$v = g \cdot t $$

and here comes into my question: how can I calculate the impact on water surface when hitting it at a certain speed? For example:

  • How much impact would I receive when hitting the water with speed 10m/s?
  • How much impact would I get when hitting the water after 3s? (Initial speed is 0)

... And that's it; I don't know if it's possible to do, but just curious.

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This is not a well defined question, although there are a few things you can do to roughly approximate an answer.

The first issue that water, and human bodies, are complicated. Water, ultimately governed by, say, the Navier stokes equation, behaves in all sorts of complicated ways. Therefore, finding the force on a human body hitting the water is also complicated. For instance, whether you are diving or belly flopping matters a lot. To really answer this question you'd need to do a computer simulation.

Now, when you ask "how much impact would I recieve" I'm going to assume by 'impact' you mean 'force.'

In order to get a rough estimate, let's make the following approximation: when the falling object is fully submerged its velocity is $0$. This is close to the truth for someone hitting the water at fast velocities not in a diving position. Let's also assume that a human being is a cube with mass $m$ and side length $L$.

When the bottom of the cube is just barely touching the water, i.e. its center of mass's height above the water is $L/2$, the cube is travelling at its free fall velocity $v$. When the cube is fully submerged, i.e. its center of mass height above the water is $-L/2$, and the cube has a velocity of $0$. Therefore, after travelling a distance of $L$, its velocity has reduced from $v$ to $0$. From the equation

\begin{align} v_f^2 - v_i^2 &= 2 a \Delta y \\ - v^2 &= -2 a L \end{align} where here we assume the cube undergoes constant acceleration, we see that $$ F = ma = m \frac{v^2}{2L}. $$

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