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Falling Object

  1. Apr 10, 2012 #1
    A helicopter is ascending vertically with a velocity of 12.5ms^-1. At a height of 120m above the ground a package is dropped out the door.

    How long does it take for the package to reach the ground ?.


    The attempt at a solution

    I know

    U = 12.5ms
    A = -9.8ms
    V = 0ms
    S = 120m

    Used t = v-u/a

    T = 1.27 Seconds - the time it takes the object to stop ascending

    Then used s = ut + .5 at^2 to find distance traveled

    S = 7.97 meters

    Adding this to the original height of 120m to get 127.97m

    So now I have:

    U = 0
    A = 9.8ms
    S = 127.97m

    Use v^2 = u^2+2as to calculate the velocity after t

    V = 50

    Use V - U/A to get time

    T = 5.1

    Now I added the ascent and decent time to get total time before it hits the grounds

    Total Time = 6.37 Seconds


    Could someone look over this for me and tell me if I am even close to the correct answer ?.

    Any help would be greatly appreciated
     
  2. jcsd
  3. Apr 10, 2012 #2
    s=ut-0.5at^2
    S is not a distance traveled but the displacement.
    It is it's position with respect to the initial location.
     
  4. Apr 10, 2012 #3

    tiny-tim

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    hi maca_404! :smile:
    no, V is not 0, is it? :redface:
     
  5. Apr 10, 2012 #4
    I had V as the velocity after t as it is ascending I assume it would need to be a two part calculations. The first when the object is ascending and being slowed by gravity at 9.8ms till its final velocity is 0 hence the V = 0 at the beginning. Then a second calculations starting from a zero velocity and then accelerating at 9.8ms towards earth. Is this not the correct procedure and can this be done with just one equation ?.
     
  6. Apr 10, 2012 #5

    tiny-tim

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    no

    use azizlwl's :smile: standard constant acceleration equation, s= ut + 1/2 at2
     
  7. Apr 10, 2012 #6
    Thank you for the replies, I am more than a bit confused now I use this ballistic calculator online http://www.convertalot.com/ballistic_trajectory_calculator.html

    Entering a initial velocity of 12.5ms and a starting height of 120m and a angle of 90Deg straight up the resulting answer is the same as I obtained originally.

    Is my answer incorrect ?. Or am I just going about it the wrong way ?

    Thanks Again
     
  8. Apr 10, 2012 #7

    tiny-tim

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    i didn't check your original answer, as soon as i saw that you'd started the wrong way i stopped reading

    your answer looks correct, but you won't get many marks in the exam for doing all that unnecessary work :redface:

    try it the short way
     
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