Time taken for free falling objects to travel a set distance

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Discussion Overview

The discussion revolves around the time taken for free-falling objects to travel a set distance on Earth, specifically focusing on the derivation of the equation t = √(2d/g) and its relation to kinematics. Participants explore the underlying principles and equations related to constant acceleration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant cites the equation t = √(2d/g) from Wikipedia and questions why the distance is doubled in this context.
  • Another participant suggests familiarity with standard kinematics equations for constant acceleration.
  • There is a repeated mention of the kinetic energy formula K = 1/2mv², which is challenged as not being a good answer to the original question.
  • Some participants express uncertainty about their knowledge of the relevant physics concepts and equations.
  • A participant indicates an interest in orbital mechanics and admits to struggling with deriving the formula from their own understanding.
  • Discussion includes the equation s = 1/2at², with participants questioning its derivation and relevance to the original equation.
  • One participant suggests using online resources to find more information about SUVAT and kinematics.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the derivation of the equation or the relevance of the kinetic energy formula. Multiple viewpoints and levels of understanding are present, indicating an unresolved discussion.

Contextual Notes

Some participants express uncertainty about their prior knowledge of kinematics and the derivation of the equations discussed. There is a reliance on external resources for clarification, and assumptions about familiarity with certain concepts are not universally held.

Jude Caird
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The time taken for a free-falling object to fall a set distance (on earth) is, according to Wikipedia, described by the equation:

t = √(2d/g).

Why is the distance doubled in this equation/ how is this derived?
 
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Do you know any of the standard kinematics equations for a constant acceleration?
 
K = 1/2mv2?
 
Jude Caird said:
K = 1/2mv2?

That's not a good answer!
 
Sorry for my ignorance.
 
Jude Caird said:
Sorry for my ignorance.

Have you just picked up this formula from Wikipedia without any prior knowledge or are you learning some physics/kinematics?
 
I am interested in orbital mechanics etc. I have a suitable understanding of physics and kinematics and the formulae associated with gravity. I tried to derive the formula from my own knowledge, but failed.
 
Jude Caird said:
I am interested in orbital mechanics etc. I have a suitable understanding of physics and kinematics and the formulae associated with gravity. I tried to derive the formula from my own knowledge, but failed.

You've never seen the equation ##s = \frac{1}{2}at^2## then?
 
PeroK said:
You've never seen the equation ##s = \frac{1}{2}at^2## then?
no
 
  • #10
PeroK said:
You've never seen the equation ##s = \frac{1}{2}at^2## then?
Is that what it's derived from?
 
  • #11
Jude Caird said:
Is that what it's derived from?

Yes, you could google for SUVAT or "kinematics". Or, try this:

http://www.examsolutions.net/maths-revision/mechanics/kinematics/suvat/derivation/tutorial-1.php
 
Last edited by a moderator:
  • #12
PeroK said:
Yes, you could google for SUVAT or "kinematics". Or, try this:

http://www.examsolutions.net/maths-revision/mechanics/kinematics/suvat/derivation/tutorial-1.php
Thank You
 
Last edited by a moderator:

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