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Constant acceleration.

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A hot air balloon ascending at a rate of 10m/s drops a package when the balloon reaches a height of 73 m. Find time to ground. Find velocity when package hits ground.


    The balloon and package, in this case, can be treated as a point particles and as such can be manipluated with genral kinematics equations.

    I apologize for my inability to display mathematical symbols when creating equations....


    2. Relevant equations



    X = X(initial) + V(initial)t + 1/2at^2

    V(final)^2 = V(initial)^2 + at


    3. The attempt at a solution


    I simply used the quadratic formula with the first equation to solve for t because the equation is already in terms of t. This produced:

    -10(+-)sq.rt(100+1460) / -10

    where:
    X=0
    X(initial)=73
    V(initial)= 10m/s
    a= -10m/s


    I chose the negative value of b^2 - 4ac to produce:

    t = -49.49/-10 = 4.949s

    Secondly, I used this time in the V(final) = V(initial) + at equation:

    V(final) = 10 + (-10*4.949) = -39.49m/s


    Am I thinking about this scenario correctly enough so that I am using the appropriate equations?? The answers, I am told, are incorrect.
     
  2. jcsd
  3. May 23, 2009 #2

    Doc Al

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    Staff: Mentor

    I don't see anything wrong with your method or answers. Why do you think they are incorrect? (Are you allowed to use g = 10, instead of 9.8 m/s^2?)
     
  4. May 23, 2009 #3
    sorry if i'm being stupid, but i think this equation is wrong.
    shouldn't it ought to be
    V(final) = V(initial) + at?


    I get the same t and V(final), so yeah. it doesn't matter.
     
  5. May 23, 2009 #4

    Doc Al

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    Staff: Mentor

    You're correct: The original post did list an incorrect equation. But the correct version was used to solve the problem.
     
  6. May 23, 2009 #5
    I am extremely sorry for the error!!

    In laboriously typing the equation out I inadvertently mixed the

    V(final)^2 = V(initial)^2 + 2ax equation with the V(final) = V(initial) +at equation.


    very sorry!! I see that I also got my units wrong for a or, in this case, g! ouch.

    thanks for the help.

    I'm sure I am able to use constant g = -10m/s^2.

    The homework is posted online through a webassign account and since it is my first experience using it, I hope my unfamiliarity is causing the confusion, not my inaccurate computations! lol
     
  7. May 23, 2009 #6

    Doc Al

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    Staff: Mentor

    Since you are familiar with this kinematic equation, why not use it to check your answer for the final speed without using the time.
     
  8. May 23, 2009 #7
    V(final)^2 = V(initial)^2 + 2ax:

    Vf^2 = 10^2 + (2)(-10)(-73)

    Vf^2 = 100 + 1460

    Vf^2 = 1560

    Vf = 39.49 m/s

    should this vector quantity be made negative to indicate downward motion?
     
  9. May 23, 2009 #8

    Doc Al

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    Staff: Mentor

    Yes. Or you can just specify the direction as downward.

    Whenever you take a square root you get two answers: one +, one -. It's up to you to choose the one that fits the situation. In this case, you know it's moving downward, thus you'd choose the negative answer (using the usual sign convention).
     
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