4/17/2023 0 Comments Freefall position vs. timeWhat you do know, or can make assumptions about, is the terminal velocity in a particular body position, for a typical skydiver. Assuming the skydiver is stable in only one position, then you can use standard cals for a falling object ( ). In this type of jump, there is only one initial force, gravity in the vertical direction, and the velocity is zero, so the drag force is zero. And like I mentioned, it might not be a whole lot different than a skydive out of a moving aircraft. If you were to look at a balloon jump, accounting for the velocity with respect to time is easier. I'm not really sure how much that is, but it might be negligible for your purposes. During this period, depending on the orientation of the skydiver, there can be some amount of lift generated from the horizontal component of the skydiver's velocity, which can complicate the calc. This transition period is known as "the hill". And finally there is the force of gravity. There is drag force in the direction of the relative wind (which begins roughly parallel to the ground and transitions to roughly perpendicular, and aligned with gravity). In the beginning of the skydive the jumper has a linear momentum in the direction of the aircraft's flight. Even then, your results won't be exactly accurate for the beginning of the skydive due to the forward speed of the aircraft, unless you're talking about a balloon jump, and then that would be easier to calc. You would have to break it down into at least two calcs. There's not an individual formula that I'm aware of that will solve for speed or height as a function of time for a whole skydive.
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