Let's imagine this in an situation where a person jumps down from a smaller building, a garage for example. The exact numbers aren't really important here, since I'm not aiming for an exact answer, but we can call it 2 metres in height.

To have both situations as similar to each other as possible, let's define them like this:

  1. In first "try", the person runs on the building, the jumps down. He is not jumping up to increase his jumping distance, he simply 'drops down', but considering that he's running with a certain speed forward, he will be falling in about 45 degrees angle. He drops at the distance X from the building.

  2. In second "try", the person runs on the building, but does a forward somersault during the flight. As before, he doesn't necessarily jumps up, as the building height is already enough for him to complete the sault and as with the previous situation, he drops at the same distanes X from the building.

In which scenario does this person hits the ground at the lower fall speed?


2 Answers 2


The person hits the ground at the same speed in both scenarios.

Once you're in the air, you fall towards the ground with a constant acceleration of about 10 m/s^2. Everything falls the same way - rocks, cannonballs, people - regardless of size or somersaulting.

There may be some small effects from air resistance, but not enough to be noticeable. The sommersaulter may also land with a different orientation so that his center of mass falls a further distance. Then he could hit the ground with slightly higher speed. But basically, everything falls the same way.

Check out this Youtube video of a feather and hammer falling simultaneously on the moon, which shows that heavy and light objects fall the same way absent air.

Also see this Youtube video of the TV show Mythbusters dropped a bullet straight down and firing one from a gun. The bullets fall in the same amount of time, regardless of their horizontal speed.


There will be points on a rotating body moving faster and slower than the CoM rate, however I assume your gymnast has pulled out of the spin and lands on their feet, then they land at the same rate, as already mentioned.

  • $\begingroup$ Can you please provide a full answer instead of a snippet? $\endgroup$
    – Sklivvz
    Commented Apr 9, 2011 at 16:20

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