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A passenger in a helicopter traveling upwards at 20 m/s accidentally drops a package out the window. If it takes 15 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?

  • physics -

    Neglect air friction.

    Solve this equation for height, H:

    Y (height above ground)
    = H + 20t -4.9 t^2 = 0

    using t = 15 s.

  • physics -

    After leaving the helicopter the package is moving upwards at the initial velocity v0.
    This is decelerated motion with acceleration –g (g≈10m/s^2)
    h0=v0t1-(gt^2)/2
    v=0=v0-gt1

    t1=v0/g=20/10=2 s.

    After this time interval the package is at the height
    h0=v0^2/2g= 20 m.

    Then the package is falling down during t2=t-t0=15-2=13 s:

    H= h0+h=(gt2^2)/2=(10x169)/2=845 m.

    Hence, the height of helicopter when the package was dropped is

    h=H-h0=845-20=825 m

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