Which Hits First? | Physics Explanation

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Homework Help Overview

The discussion revolves around the physics of falling objects, specifically comparing the motion of two stones dropped from different heights and whether they hit the ground simultaneously. The subject area includes concepts from kinematics and gravitational acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between displacement and time for falling objects, questioning the validity of the original poster's equation. Some participants emphasize that all objects fall at the same rate in a vacuum, while others point out that height differences affect fall times.

Discussion Status

The discussion is active, with various perspectives being shared. Some participants affirm the equation's validity, while others critique its usefulness in understanding the physical situation. There is acknowledgment of differing interpretations regarding the effects of height and external factors like friction.

Contextual Notes

Participants note the importance of considering external factors such as inclined planes and wind resistance, which may not have been fully addressed in the original question. The original poster expresses a desire for clarity and understanding, indicating a learning context.

Misr
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Hello People peace be upon you
http://img693.imageshack.us/img693/4447/whichhitsfirst.jpg
I have an explanation for this

From the equation of motion :

X = V0t + 1/2 a t^2
Where X is the displacement covered by the body
Let the displacement covered by the first stone X1 and the displacement covered by the second ball X2

To fall at the same time the two displacements must be equal right??
Therefore X1=X2
Therefore V0t + 1/2 g t1^2 = V0t + 1/2 g t2^2
Since the initial velocity is zero in both cases
Therefore V0t = zero
And the acceleration due to gravity (g) is constant for any bodies falling freely
Therefore 1/2* g* t1^2 =1/2 *g*t2^2
By dividing the equation by 1/2 g
Therefore t1^2 =t2^2
Therefore t1=t2

IS THAT RIGHT??

Thanks
 
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you don't even need to do calculations. We know that gravity near the Earth's surface is 9.8m/s² of acceleration. So that means all objects, no matter a feather or a stone or a brick all fall at the same speed if you do it in a vacuum. The only thing that can change this is wind resistance.
 
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your equation is trivial by the way.
 
Up >>
 
I forgot to add a point
when we through two objects from diffrent heights they don't fall at the same time
although they are facing the same gravitational acceleration this is the idea of my equation I will repeat my Question :DOES THE EQUATION WORKS?

Thanks people
 
dacruick said:
your equation is trivial by the way.
Maybe for people who have advanced past introductory physics. Not so for many who are learning it for the first time.

Misr said:
I forgot to add a point
when we through two objects from diffrent heights they don't fall at the same time
although they are facing the same gravitational acceleration this is the idea of my equation I will repeat my Question :DOES THE EQUATION WORKS?

Thanks people
Yes, the equation works.
 
Yes, the equation works.
Thanks so much
 
The equation is trivial no matter from which point you look at it. You have a vacuum situation. I don't mean to be confrontational RedBelly, but I don't believe that is the way to help someone understand the concept.

The equation is trivial because every single one of your variables is the same. Thats the same as saying v*x*y = x*v*y. All I am doing is cancelling both sides and getting v=v. The equation "works", but is of no use, in getting any correct answer, or in understanding anything about the physical situation
 
The OP did distinguish between x1 and x2, as well as t1 and t2.

I agree that your explanation in Post #2 is conceptually better.
 
  • #10
To the OP: you didn't take into consideration the inclined plane. You also didn't consider friction. I think the question wanted you to consider both.
 
  • #11
calm down calm down people we r not fighting and i do agree with all your opinions I am very new to physics and ur trying to help so thanks 4 all of u and no use 4 getting so nervous I am just learning.actually mr dacruik i already know what u said but u dun even put the displacdment in ur consideration this what made me think of this trivial eqn as u said thnx anyways.
ideasrule;thanks but i was talkin bout the 1st case plus i don't know how
 
  • #12
Misr said:
calm down calm down people we r not fighting and i do agree with all your opinions I am very new to physics and ur trying to help so thanks 4 all of u and no use 4 getting so nervous I am just learning.actually mr dacruik i already know what u said but u dun even put the displacdment in ur consideration this what made me think of this trivial eqn as u said thnx anyways.
ideasrule;thanks but i was talkin bout the 1st case plus i don't know how
Goodness! I think Misr has had a stroke! His writing, initially excellent, has slurred into unintelligibility...
 
  • #13
Goodness! I think Misr has had a stroke! His writing, initially excellent, has slurred into unintelligibility...
Thanks for the help
 

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