Which Distance is Correct for a Free Falling Object?

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Discussion Overview

The discussion revolves around the calculation of distance traveled by a free-falling object under the influence of gravity, specifically comparing two different formulas for distance: one assuming constant velocity and the other accounting for acceleration. The scope includes conceptual understanding and mathematical reasoning related to kinematics.

Discussion Character

  • Conceptual clarification
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents two formulas for calculating distance: one based on constant velocity (d = vt) and another based on acceleration (d = 1/2 a*t^2).
  • Another participant corrects the first by stating that the velocity of a free-falling object is not constant, emphasizing that the acceleration is 9.8 m/s².
  • A participant clarifies that when referring to v = 9.8 m/s, they mean the instantaneous velocity at the end of the first second.
  • Another participant argues that while the instantaneous velocity at the end of the first second is 9.8 m/s, the average speed during that second is actually 4.9 m/s due to starting from rest.

Areas of Agreement / Disagreement

Participants express disagreement regarding the interpretation of distance traveled in the first second of free fall, with some asserting that the distance is 9.8 m and others arguing it is 4.9 m based on average speed. The discussion remains unresolved.

Contextual Notes

There is a lack of consensus on the interpretation of instantaneous versus average velocity in the context of free fall, and the assumptions regarding constant velocity versus acceleration are not fully reconciled.

arunkumarg
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I have a simple question may sound stupid for most of you:

For a free falling object v=9.8 m/sec. which of the distance traveled per sec is correct:

Formula 1:

d = vt
d = 9.8/1 = 9.8 m

Formula 2:
d = 1/2 a*t^2

acc of a free falling object is 9.8m/sec^2

d = 1/2 9.8*1^2
d = 4.9 m

This is a bit confusing.
 
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arunkumarg said:
For a free falling object v=9.8 m/sec.
No, the acceleration of a free falling object is 9.8 m/s^2. The velocity is not constant.

Formula 1:

d = vt
d = 9.8/1 = 9.8 m
This is incorrect as it assumes a constant velocity.

Formula 2:
d = 1/2 a*t^2

acc of a free falling object is 9.8m/sec^2

d = 1/2 9.8*1^2
d = 4.9 m
This is correct. That's the distance fallen in the first second.
 


Thanks for your reply.

When I say V = 9.8m/sec. its the instantaneous velocity (for the 1st Sec).

Hence, for the 1st sec the dist must be 9.8m.

I know I am missing the point somewhere.
 


arunkumarg said:
When I say V = 9.8m/sec. its the instantaneous velocity (for the 1st Sec).
That's the instantaneous velocity at the end of the 1st second. It starts out with zero velocity at t = 0 and ends up going 9.8 m/s at t = 1 sec.
Hence, for the 1st sec the dist must be 9.8m.
That would be true if the speed were constant for the entire second. But it's not. Since it starts from rest, the average speed during the 1st second is only 4.9 m/s.
 


Sounds good for me... Thanks pal
 

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