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I'm a high school physics student, and I've been puzzling over how one can derive the equations regarding projectile motion geometrically. To do so, I began thinking of projectile motion in terms of circular motion, for 2 main reasons, including:

  • All projectile motion problems in high school physics involve perfectly symmetric parabolas, and can thus be interpreted to be part of a circle; a circular segment.
  • At just the right theta & resultant velocity, a projectile can be launched such that begins orbiting the earth, or whatever planetary body the problem concerns, thus resulting in a form of circular motion

Exactly why may this be useful? Well, the vertical & horizontal displacement could be found much easier (in some cases), using simple geometry.

Question: Exactly how can one derive the equations for projectile motion geometrically, via the same means we approach circular motion problems?

Diagram provided below: Circular segments

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  • $\begingroup$ "Well, the vertical & horizontal displacement could be found much easier (in some cases), using simple geometry." What cases are you thinking about? Projectile motion is parabolic, which is NOT easily handled with circles. $\endgroup$ – Bill N Dec 9 '17 at 4:30
  • $\begingroup$ "high school physics [problems] involve perfectly symmetric parabolas", typically without air resistance. $\endgroup$ – DarkRunner Dec 9 '17 at 4:36
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    $\begingroup$ Your first bullet point is incorrect. A parabola is a conic section not part of a circle; a circular segment. This invalidates the rest of your reasoning. $\endgroup$ – John Rennie Dec 9 '17 at 5:32
  • $\begingroup$ Most high school physics problems deal with parabolas in a two dimensional space, which is why I believe it would be much easier to analyze it as a segment of a circle. $\endgroup$ – DarkRunner Dec 9 '17 at 15:33
  • $\begingroup$ @DarkRunner Parabolas in two dimensional space are not circular segments though. They have different shapes. Why use something wrong when we can use the easier parabola? $\endgroup$ – JMac Dec 9 '17 at 18:54
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" Exactly why may this be useful? Well, the vertical & horizontal displacement could be found much easier (in some cases), using simple geometry. "

Well,in most of the high school problems,the only parameters given to you are angle theta and initial velocity u. I don't get your point of calculating displacement easily.

Through my understanding of your question, basically you want to say that we can consider parabola as a segment of circle.

Well,that would not be completely wrong but I want to add that a segment of circle too is only roughly parabolic.

According to me,it would be tougher actually.....

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  • $\begingroup$ "Well,that would not be completely wrong." It would be completely wrong. Most parabolas are not even close to a circular arc. $\endgroup$ – Chris Dec 9 '17 at 22:22

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