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

AI Thread Summary
The discussion focuses on the relationship between the position (S) of a freely falling body and the square of time (t^2), highlighting that the slope of the S vs. t^2 graph equals g/2 due to the kinematic equation s = 0.5 * g * t^2. This equation indicates that the distance fallen is proportional to the square of the time, which results in a linear relationship when graphed. The slope represents half of the acceleration due to gravity (g), explaining why it is g/2 rather than g itself. Participants confirm their understanding of this relationship and the underlying physics concepts. The discussion reinforces the connection between kinematic equations and graphical representations of motion.
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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