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Finding the time when two stones cross paths

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.69 m. The stones are thrown with the same speed of 9.97 m/s. After what time do the stones cross paths?

    2. Relevant equations
    d = vt + 0.5at2

    3. The attempt at a solution
    distance travelled by both rock A and rock B combined is 6.69
    DA + DB = 6.69

    let the upwards direction be positive

    DA: (top of cliff)

    d = vit + 0.5at2
    = (-9.97 m/s[up])t + 0.5(-9.8 m/s2[up])t2
    = -9.97t - 4.9t2

    DB: (base of cliff)
    d = vit + 0.5at2
    d = (9.97 m/s [up])t + 0.5(-9.8m/s2[up])t2
    d = 9.97t - 4.9t2

    DA + DB = 6.69
    -9.97t - 4.9t2 + 9.97t - 4.9t2 = 6.69
    -9.8t2 = 6.69 getting negative time...?
    t = 0.82 s

    the correct answer is 0.335 seconds. I think I am messing up my directions and signs for my vectors... any help?
     
  2. jcsd
  3. Feb 18, 2014 #2
    Good start.
    Three things…

    1. Keep your numbers as variables. It's alot easier to keep writing "g" than to keep writing "-9.8 m/s2[up]".
    2. You said.

    "distance travelled by both rock A and rock B combined is 6.69
    DA + DB = 6.69"

    THis is close, but not quite correct. We don't know that this is the case. It would work if A was going down the whole time and B was going up the whole time. We know A will be going down the whole time, but B will reach a maximum height and begin to fall.

    Your equation should mathematically say " The height of the cliff is 6.69 m".
     
  4. Feb 18, 2014 #3
    Missed my third point…

    "-9.8t2 = 6.69 getting negative time…?"

    Nope, you wouldn't get negative time. It's even weirder; You would have imaginary time!
     
  5. Feb 18, 2014 #4

    Dick

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    Homework Helper

    It's probably easier to think about if you pick a common origin for ##D_A## and ##D_B##. Pick the common origin ##D=0## at the base of the cliff. So ##D_A=-9.97t - 4.9t^2 + 6.69## and ##D_B=9.97t-4.9t^2##. Now just solve ##D_A=D_B##.
     
  6. Feb 19, 2014 #5
    ##D_A + 6.69 = D_B##

    It took me a while, but I finally found meaning behind this equation...
    displacement of rock A: ##D_A = -6.69 m [up]##, and B: ##D_B = 0 m.##

    -6.69 + 6.69 = 0, brilliant!
     
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