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Hiker in the wilderness

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A hiker has managed to strand himself in the wilderness. A rescue plane traveling at 80.0 km/h horizontally and at an altitude of 70.0 m airdrops some survival supplies for his use. If the goods are thrown down at 2.50 m/s in the vertical direction with respect to the plane, how far in advance of the person must the drop occur so that the gear lands directly at his feet? Ignore air resistance.

    2. Relevant equations
    80 km/h= 22.2 m/s
    xf= xi + vit + (1/2)at^2

    3. The attempt at a solution
  2. jcsd
  3. Sep 24, 2009 #2
    if it goes down with constant speed 2.5m/s it is:

    70/2.5 = 28 s

    28 * 22.2 = 621.6
  4. Sep 24, 2009 #3
    ^ uh no, there's gravity.

    The object is moving forward at 22m/s forward and 2.5m/s downward initially. So just find out how long it will take to reach the ground when dropped from 70m. Then you can find the distance it will travel in the x-dir.
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