Understanding Freely Falling Bodies: The Relationship Between S and t^2

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Homework Help Overview

The discussion revolves around the relationship between displacement (S) and time squared (t^2) for a freely falling body under the influence of gravity. Participants are exploring the implications of the slope of the S vs. t^2 graph and its relation to gravitational acceleration (g).

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions why the slope of the S vs. t^2 graph yields g/2 instead of g, and whether the graph demonstrates a linear relationship. Other participants reference kinematic equations to support their reasoning.

Discussion Status

The conversation includes attempts to clarify the relationship between displacement and time squared, with some participants affirming their understanding of the kinematic equation. However, there is no explicit consensus on the implications of the slope.

Contextual Notes

Participants are discussing the scenario of a freely falling body with zero initial velocity, which may limit the context of their analysis.

ritwik06
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1.For a freely falling body under gravity:
1.Why does the slope of S vs. t^2 gives g/2? Why not g itself?
2.Does the graph exhibit linear relationship?

I have studied that the slope of a velocity time graphy gives the acceleration directly. Now when we make a graph for S vs. t^2 from a graph of velocity-time for a freely falling body. So why does the slope of the graph obtained gives g/2?
 
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Think of the equation relating position and acceleration (kinematic equation) for a particle with zero initial velocity
 
cristo said:
Think of the equation relating position and acceleration (kinematic equation) for a particle with zero initial velocity

s=0.5*gt^2

right??
 
I got it! Thanks!
 

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