1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where When Physics Problem with 1 obj going up and the other going down.

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    An object is thrown downward with an initial speed of 10 m/s from a height of 60m above the ground. At the same instant, a second object is propelled vertically upward from ground level with an initial speed of 40 m/s. At what height above the ground will the two objects pass each other?


    2. Relevant equations
    Professor said for where+when problems should always use the formula
    Xf = X0 + (VX0)(Δt) + (1/2)(AX0)(Δt)^2

    Variable key:
    Xf = final x
    X0 = initial x
    VX0 = initial velocity
    AX0 = acceleration due to gravity (-9.8m/s^2)
    Δt = change in time


    3. The attempt at a solution
    Ok so first I mapped out the known factors in the equation for each motion of interest
    For motion A we have:
    X0 = 60, VX0 = 10, AX0 = 9.8, missing Δt and Xf
    For motion B we have:
    X0 = 0, VX0 = 40, AX0 = 9.8, again missing Δt and Xf

    I figure that we are looking for the X when both x's are zero so I set them to equal each other in the same form as the equation
    AX0 + AVX0(Δt) + (1/2)(AX0)(Δt)^2 = BX0 + BVX0(Δt) + (1/2)(AX0)(Δt)^2

    60 + 10t + 4.9t^2 = 40t - 4.9t^2

    From this I get everything to one side so the right side is 0 and I can use the quadratic formula to solve for Δt

    Equation looks like
    9.8t^2 - 30t + 60 = 0

    So i do the quadratic equation
    30 +/- sqrt(30^2 - 4(9.8)(60)) all over 2(9.8)

    I end up with 3.47475 OR -.413526 as Δt.

    I plug either one of those numbers into either one of the original equations to solve for Xf and I get numbers that are no where near correct.

    He has listed the answer as 41 m.
    Am I going about this completely wrong?

    Thanks,
    Cory
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2012 #2
    I have not looked at your work too closely, but I will say right off the bat, I think you might have some sign issues. You need to choose what direction you will consider positive and what you will consider negative. Your initial velocities are in opposite directions and so they must have opposite signs. The values of acceleration will have opposite signs as well.
     
  4. Jan 30, 2012 #3
    Thank you I was wondering if that might have been my issue. I flipped the signs on acceleration but I guess all of them have to go in the opposite directions. I will retry and post my work here. Thank you!

    EDIT: Tried changing the sign on one of the velocities... came out with wrong answers... Then tried changing the other motion's sign instead and it was wrong again.
    Gonna try again tomorrow... been at this for hours on 1 problem lol.

    Thanks!
     
    Last edited: Jan 30, 2012
  5. Jan 30, 2012 #4
    OK I lied ... I stayed up and tried it again... I have a value that is close for distance... 45.04308602 meters(all values kept in calculator memory so not rounded)... His answer is 41m I wonder if it could just be a difference in rounding or something... i'm not sure :(

    When I plug my time that i calculated from the quad formula into the original 2 distance equations:
    Xf = 60 - 10(t) - 4.9t^2
    and
    Xf = 40t + 4.9t^2

    They both give me the same answer:
    Xf = 45.04308602 ... Does that mean that it IS the distance where they cross paths?

    Thanks.
    Cory
     
  6. Jan 31, 2012 #5
    Bump: Can anyone tell me if the methodology used to check my answer is correct?
    It seems like it would make sense but I don't like that my answer doesn't agree with the professor's.
     
  7. Jan 31, 2012 #6
    I got 41 meters. You need to still be careful with your signs. Look at the following:

    The term with 4.9 in it is your 'gravity term.' If you choose 'up' to be positive (which you have, by putting a (-) sign in front of the 10t in the 1st equation) then since gravity points downward, the gravity terms must be negative right? So why is is the gravity term negative in the 1st equation and positive in the 2nd ? Gravity is pointing upward in for object B? :wink:
     
  8. Feb 1, 2012 #7
    Thank you very much I finally figured it out.

    So in the equation when I always have gravity as negative it works.
    I think I had tried it before that way but in writing a step of the equation I had left numbers off or changed signs... It would seem I need to be extremely careful with making sure I write the initial equation right and every subsequent operation because 1 tiny mistake will make me scratch my head for days :D.

    Thanks again Saladsamurai.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Where When Physics Problem with 1 obj going up and the other going down.
Loading...