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Maximum height

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A passenger in a helicopter traveling upwards at 17m/s accidentally drops a package out the window.
    a)If it takes 13s to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?
    b)what was the maximum height of the package above the ground?



    3. The attempt at a solution

    a)
    V=17 m/s
    t=13s
    g=-9.8m/s^2

    so find x-x0

    by using X-Xo=(1/2 a t^2) + Vo t

    i got 607. which i am sure is the right answer

    b)i know maximum height is when V(final)=0

    so i did 17m/s=(9.8m/s^2) t + 0

    and i ended up with t=1.73

    plugging this back into the equation i get the extra amount of distance
    (X-Xo)= (1/2)(-9.8)(1.73^2)+(17)(1.73)

    and that gives me x=14.7

    So for the maximum height i figure u just add 14.7+607= 621.7

    However the book is telling me this answer is wrong. Correct answer is 627

    Did i do something wrong??
     
  2. jcsd
  3. Jan 16, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your solution looks good to me.
     
  4. Jan 17, 2009 #3
    I got 607 (607.1) m and 622 (621.8) m so I guess you are right. Plus, 7<=>2 is a common typo.
     
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