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Which hits First

  1. Feb 10, 2010 #1
    Hello People peace be upon you
    http://img693.imageshack.us/img693/4447/whichhitsfirst.jpg [Broken]
    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
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 10, 2010 #2
    you dont 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.
     
    Last edited by a moderator: Feb 11, 2010
  4. Feb 10, 2010 #3
    your equation is trivial by the way.
     
  5. Feb 11, 2010 #4
    Up >>
     
  6. Feb 11, 2010 #5
    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
     
  7. Feb 12, 2010 #6

    Redbelly98

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    Maybe for people who have advanced past introductory physics. Not so for many who are learning it for the first time.

    Yes, the equation works.
     
  8. Feb 12, 2010 #7
    Thanks so much
     
  9. Feb 12, 2010 #8
    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 im 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
     
  10. Feb 12, 2010 #9

    Redbelly98

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    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.
     
  11. Feb 12, 2010 #10

    ideasrule

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    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.
     
  12. Feb 12, 2010 #11
    calm down calm down people we r not fighting and i do agree with all your opinions im very new to physics and ur trying to help so thanx 4 all of u and no use 4 getting so nervous im 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 dunno how
     
  13. Feb 12, 2010 #12

    DaveC426913

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    Goodness! I think Misr has had a stroke! His writing, initially excellent, has slurred into unintelligibility...
     
  14. Feb 13, 2010 #13
    Thanks for the help
     
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