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Homework Help: Relative motion (2 arrow shot up find meeting distance)

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Two arrows are fired straight up. The first is at 40 m/s and the second at 60 m/s and 3 seconds after arrow one. At what hieght will the arrows meet in the air.

    2. Relevant equations
    ?
    s=ut+1/2 at^2

    3. The attempt at a solution

    s1=s2
    t2=t1+3
    u1(t2+3)+1/2a(t2+3)^2=u2t2+1/2at2^2
     
  2. jcsd
  3. Jan 26, 2010 #2

    rl.bhat

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    Homework Helper

    Hi GregR, welcome to PF.
    Your approach is correct.
    But
    t2=t1+3 it should be t1=t2+3.
    In the equation you have substituted correctly.
    Now proceed.
     
  4. Jan 26, 2010 #3
    So then:
    u1t2+3u1+.5at22+4.5a=u2t2+.5at22

    u1=40m/s
    u2=60m/s
    a=-9.81m/s2

    The .5at22 on each side should cancel each other out leaving:

    u1t2+3u1+4.5a=u2t2

    Solving:

    40t2+3(40)+4.5(-9.81)=60t2
    40t2+120-44.145=60t2
    20t2=75.855
    t2=75.855/20
    t2=3.79275 seconds
     
  5. Jan 26, 2010 #4
    If I use this as the time and recalculate the distance for each arrow it doesn't jive.

    Through trial and error I found the meeting time to be 1.5345 seconds at a distance of 80.52 meters from the ground.

    Not sure if I'm missing something or its just a minor error on my part.
     
  6. Jan 26, 2010 #5

    rl.bhat

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    Your calculation is wrong.
    The equation will be
    u1(t+3) - 0.5*g*(t+3)^2 = u2*t - 0.5*g*t^2
    Simplify the equation and solve for t.
     
  7. Jan 26, 2010 #6
    Using this formula:
    u1(t+3) - 0.5*g*(t+3)^2 = u2*t - 0.5*g*t^2

    From this I simplified as follows:

    u1t +3u1-0.5*9.81*(t2+9) = u2*t - 0.5*9.81*t2

    u1t +3u1-4.905*(t2+9) = u2*t - 4.905*t2

    u1t +3u1-4.905*t2-4.905*9 = u2*t - 4.905*t2

    The -4.905*t2 on both sides should cancel out leaving.

    u1t +3u1-44.145= u2*t

    Inserting u1=40m/s and
    u2=60m/s

    40*t + 30*40 - 44.145 = 60*t
    40*t + 120 - 44.145 = 60*t
    75.855 = 60*t - 40*t
    20*t = 75.855
    t = 75.855 / 20
    t = 3.79275

    I wrote it up step by step and got the same results.
    Not sure where I'm going wrong here.
    Please let me know.
    Thanks
     
  8. Jan 27, 2010 #7

    rl.bhat

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    (t+3)^2 = t^2 + 6*t + 9
     
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