Solve Gravity Lab Problem: Determine Accel Due to Gravity

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SUMMARY

The discussion focuses on determining the acceleration due to gravity from a quadratic fit equation, specifically y=4.97x^2+1.53x+0.0503. Participants confirm that the coefficient 4.97 represents half of the gravitational acceleration, leading to a calculated value of g as 9.94 m/s². The initial velocity (Vnaught) is identified as 1.53 m/s. The conversation clarifies the correct interpretation of the quadratic equation in the context of motion under gravity.

PREREQUISITES
  • Understanding of quadratic equations in physics
  • Familiarity with kinematic equations, specifically y=1/2gt^2+Vnaught(t)+ynaught
  • Basic knowledge of gravitational acceleration and its standard value
  • Ability to interpret coefficients in polynomial equations
NEXT STEPS
  • Study the derivation of kinematic equations in physics
  • Learn how to perform quadratic regression analysis
  • Explore the effects of initial velocity on projectile motion
  • Research the differences between theoretical and experimental values of gravitational acceleration
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Students in physics, educators teaching kinematics, and anyone interested in understanding the application of quadratic equations in motion analysis.

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Homework Statement


Suppose that a quadratic fit to a position plot yields the following fit result: y=4.97x^2+1.53x+.0503, how would I determine the accelearation due to gravity?





Homework Equations


I know that y=1/2gt^2+Vnaught(t)+ynaught. I know that g is 9.87m/s^2. So in my original given equation, I know that 4.97 is half of the gravity, and 1.53 is my Vnaught. Would this be the answer they are looking for, or is there something I am missing?


The Attempt at a Solution


I may be exploring this question deeper then I should. I believe that the acceleration due to gravity is 9.8m/s^2, and the initial velocity of the particle would be 1.53m/s?
 
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Nope, you have the right idea (though I for one wouldn't have written the initial equation as y = 4.97x^2+1.53x+.0503 - I would have written it as y=4.97t^2+1.53t+.0503, but that's just me being picky.)
 
That is just the way the lab gave the initial equation. I am working on my preliminary assignment. So I haven't actually done the problems yet.
 

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