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Let's say you are launched out of a canon at a $0^\circ$ degree angle 5 feet off of the ground. Normally, from my basic understanding of free falling, you have to be move unimpeded with no force acting upon you. If you are shot horizontally, would you be considered free falling if you are consistently moving at terminal velocity of 122mph?

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  • $\begingroup$ how is 122 mph considered "terminal velocity"? are you on an asteroid somewhere? $\endgroup$ – robert bristow-johnson May 20 at 19:19
  • $\begingroup$ How is a terminal velocity compatible with a free fall? Is there drag? $\endgroup$ – Qmechanic May 20 at 19:21
  • $\begingroup$ I don't know the answer to these questions. I Wikipedia'd some facts before asking. You could be an asteroid somewhere $\endgroup$ – Chris Zog May 20 at 19:23
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    $\begingroup$ @ChrisZog So you are saying the barrel of the canon is horizontal? $\endgroup$ – Bob D May 20 at 19:32
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    $\begingroup$ Your horizontal velocity has no effect on how fast you fall. From a height of 5 feet it will take 0.5575 seconds to hit the ground, whether you have any horizontal motion or not. $\endgroup$ – BillDOe May 20 at 20:12
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As I recall free fall means unrestricted downward motion in a gravitational field. Any air drag (and terminal velocity) you experience due to your horizontal velocity leaving the canon should have no effect on your downward acceleration $g$.

The only restriction to free fall would be air drag on your downward motion. Since your height out of the canon is only 5 feet, that's hardly enough falling distance for air drag to play any significant role, and you certainly would not reach any downward terminal velocity.

Perhaps I am misunderstanding your scenario. If that's the case let me know.

Otherwise, hope this helps.

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