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Kinematic problem

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data
    MJ falls from rest from a tall building. 1.5 seconds later SP throws himself downward with an initial velocity of -45 meters per second. Find the distance where they meet.
    variables:
    α1= -9.81 α2=-9.81
    Δγ1 = ? Δγ2=?
    Δ†1=? Δ†2=? + 1.5
    ∨i1=0 ∨i2=-45
    ∨f1=unknown ∨f2=unknown

    2. Relevant equations
    vf=vi+aΔt (distance is absent)
    Δγ=viΔt+½αΔt^2 (final velocity is absent)

    3. The attempt at a solution
    I think I have the problem right. I found the final velocity and distance where MJ would be after the 1.5 seconds. then I used that final velocity as the initial velocity for MJ. and used that distance as MJ's starting distance. then I used true statements to say the accelerations are the same in both equations, the times are the same in both equations (since I changed my initial velocity and distance) and My distance for spiderman - 11.04 (what I got for the initial distance of MJ) equals MJ's distance. then I used Δγ=viΔt+½Δt^2, and set it up as
    11.04=vi1Δt1+½αΔt1^2-vi2Δt2+½αΔt2^2. since I knew my times and accelerations were the same, I knew subtracting ½αΔt2^2 from ½αΔt1^2, leaving me with 11.04=Vi1Δt1- Vi2Δt2. I then subtracted the initial velocities because the times are the same. Then I divided 11.04 by what I got. Would this way give me the correct answer?
     
    Last edited: Oct 19, 2014
  2. jcsd
  3. Oct 19, 2014 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Man, I though the problem was talking about Michael Jordan and Scottie Pippen!

    You should remember that MJ (Michael Jordan or Mary Jane) are still falling after 1.5 sec (and accelerating, no less), so your static approach to finding the time MJ and SP are at the same distance from the top of the building requires a different approach. Try writing separate equations which describe the position of MJ and SP separately, using the fact that both start from the same point.
     
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