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Initial Speed

  1. Jun 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A hot air balloon has just lifted off and is rising at the constant rate of 1.87 m/s. One of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 11.6 m/s. If the passenger is 2.70m above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger.

    2. Relevant equations
    Not really sure, but i was able to draw a FBD not exactly sure what else to do.
  2. jcsd
  3. Jun 3, 2007 #2
    i dont understand, the initial speed of the camera is given?
  4. Jun 3, 2007 #3
    it gives an initial speed, but it wants to know the MINIMUM initial speed that the camera can be thrown.
  5. Jun 3, 2007 #4
    ok so do you know your position function?
  6. Jun 3, 2007 #5
    no, im not sure i know what a position function is...
  7. Jun 3, 2007 #6
    [tex] s = s_o + V_o + \frac{1}{2}at^2[/tex]
  8. Jun 3, 2007 #7
    using the above position functions, write out the position equations for the balloon and for the camera.

    Balloon: [tex]s_b = 2.7 + 1.87t[/tex]
    Camera: [tex]s_c = V_ot - .5*g*t^2[/tex]

    You want to find the minimum V_o such that [tex]s_b=s_c[/tex] has a solution for t > 0
  9. Jun 4, 2007 #8
    oooo you're going to get in trouble
  10. Jun 4, 2007 #9
    y does the first one (balloon) have 1.87t? shouldnt it just be 1.87, since a=0, so .5*0*t^2=0 right?

    so for Sb=4.57 is that right?
    then im stuck on Sc, i have Sc= 0 + x + 4.9

    what do i use for T?
    what do u mean?
  11. Jun 4, 2007 #10
    do i plug in T= V - Vo/a

    if so do i use 9.8 for a?

  12. Jun 4, 2007 #11
    He means that you aren't supposed to post solutions, and a moderator will delete the solution when he sees it.

    The first balloon has 1.87t because you have a non-zero velocity term in there (which is 1.87).

    How do you get that the balloon time 4.57 in the first place, and why are you not realizing that the two should be the same? Let me ask you this, if you want to spit on a car while standing on a bridge, do you find a spitting time versus a car time and then hope your spit suspends itself in time? No, you want the times to be the same so that the two motions intersect at the desired position, or target. Is this case really any different?
  13. Jun 4, 2007 #12
    You missed a term. The equation should be..

    [tex]s = s_o + V_o t + \frac{1}{2}at^2[/tex]
  14. Jun 4, 2007 #13
    ok i think i get it now.

    so its:

    right? but im still a little lost as to what to plug in for t. If i use another equation i can do t= V-Vo/a
    but what do i use for V?
    And, does a=9.8?

    Last edited: Jun 4, 2007
  15. Jun 4, 2007 #14
    ^^so is this right?
  16. Jun 4, 2007 #15
    Yes, you got it.

    Solve for t (you might want to consider the quadratic formula) and then you can figure out Vo.
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