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Sandbag is dropped from moving ballon

  1. Jan 25, 2008 #1
    nevermind, ive solved it, thanks anyway

    hello, ive tried this simple problem a few times and mastering physics keeps taking points off =(.
    ive never used latex so bare with me

    1. The problem statement, all variables and given/known data
    A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground View Figure . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.

    [​IMG]

    Compute the velocity of the sandbag at a time 1.15 s after its release.

    2. Relevant equations
    x = X[tex]_{0}[/tex] +V[tex]_{0}[/tex]T + .5aT[tex]^{2}[/tex]
    V[tex]^{2}[/tex] = V[tex]^{2}_{0}[/tex] + 2a(x - x[tex]_{0}[/tex])

    3. The attempt at a solution
    I assume that the initial velocity of the sandbag once it is released is 5ms since that was the velocity of the balloon.

    first i found the position of the sandbag at 1.15.
    x = 40 + 5(1.15) + .5(-9.81)(1.15)[tex]^{2}[/tex]
    x = 39.3

    up to this point i am 100% sure i am correct.

    next i inserted this information into the velocity formula.
    V[tex]^{2}[/tex] = V[tex]^{2}_{0}[/tex] + 2a(x - x[tex]_{0}[/tex])

    V[tex]^{2}[/tex] = (5)[tex]^{2}[/tex] + 2(-9.81)(39.3 - 40)
    [tex]\sqrt{V^{2}}[/tex] = [tex]\sqrt{38.688}[/tex]
    V = 6.22

    Mastering physics says incorrect and to check my signs,, ive checked my signs i cant find my error

    edit: ops, since its falling i take the negative root not the positive root, problem solved..answer was -6.27
     
    Last edited: Jan 25, 2008
  2. jcsd
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