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Fun, but difficult mechanics question

  1. Aug 26, 2004 #1
    Any help would be greatly appreciated.
    Thanks in advance.

    A stone is thrown vertically upward at a speed of 45.50 m/s at time t=0. A second stone is thrown upward with the same speed 3.810 seconds later.

    1) At what time are the two stones at the same height?
    2) At what height do the two stones pass each other?
    3) What is the upward speed of the second stone as they pass each other?

    I'm pretty sure I could do #2 and #3, as long as I got some help first on #1.

  2. jcsd
  3. Aug 26, 2004 #2

    h1 = v*t - .5*g*t^2
    h2 = v*(t-3.81) - .5*g*(t-3.81)^2
    if h1=h2 , then
    t = v/g + 3.81/2

    if we have g = 10.0 and v = 45.50 , we get
    t=6.455 seconds
  4. Aug 27, 2004 #3
    Here, Ill write it out fully for you (with units), since Rogerio already solved it:

    [tex] d_1 = Vt + \frac{1}{2}at^2 [/tex]

    [tex] d_2 = V(t-3.810s) + \frac{1}{2}a(t-3.810s)^2 [/tex]

    [tex] d_1 = d_2 [/tex]

    [tex] Vt + \frac{1}{2}at^2 = V(t-3.810s) + \frac{1}{2}a(t-3.810s)^2 [/tex]

    subbing in your values:

    [tex] (45.5m/s)t - (4.9m/s^2)t^2 = (45.5m/s)(t-3.810s) - 4.9m/s^2(t-3.810s)^2 [/tex]

    [tex] (45.5m/s)t - (4.9m/s^2)t^2 = ((45.5m/s)t - 173.4m) - 4.9m/s^2(t^2 - (7.62s)t + 14.5s^2) [/tex]

    [tex] -(4.9m/s^2)t^2 = - 173.4m - 4.9m/s^2(t^2) + (37.3m/s)t - 71.05m [/tex]

    [tex] 173.4m + 71.05m = (37.3m/s)t [/tex]

    [tex] 244.45m = (37.3m/s)t [/tex]

    [tex] \frac{244.45m}{37.3m/s} = t [/tex]

    [tex] t = 6.55s [/tex]
  5. Sep 6, 2005 #4
    I have this same problem. I got the first part, but I have no idea where to start for the second part. Can someone give me a hint?
  6. Sep 6, 2005 #5

    Doc Al

    User Avatar

    Staff: Mentor

    If you did the first part, then you know how to write the height as a function of time. And you now have the time. Plug it in!
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