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    mrmotl's Avatar
    mrmotl Posts: 4, Reputation: 1
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    #1

    Sep 23, 2009, 09:42 AM
    A small bag is released from a helicopter that is climbing steadily at 1.5 m/s. after
    A small bag is released from a helicopter that is climbing steadily at 1.5 m/s. after 2 seconds,

    (a) find the velocity of the mail bag?

    (b) how far is it below the helicopter?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Sep 23, 2009, 09:49 AM

    There is not enough information. When exactly was the bag released? How high, or for how much time has the helicopter been rising? If it's after a short while, the bag may hit the ground before 2 seconds...
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    mrmotl Posts: 4, Reputation: 1
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    #3

    Sep 23, 2009, 09:57 AM

    It douse not hit the ground (a) should be 20
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    mrmotl Posts: 4, Reputation: 1
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    #4

    Sep 23, 2009, 09:58 AM
    It was released the question is for 2 seconds after it was released
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    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Sep 23, 2009, 10:03 AM

    Ok, use the formula

    v is the speed of the bag
    u the velocity of the helicopter when the bag was released
    a the acceleration due to gravity
    t the time elapsed.
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    mrmotl Posts: 4, Reputation: 1
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    #6

    Sep 23, 2009, 10:10 AM

    So would (a) be 21.5 m/s?
    Do u know (b) or how would I get it
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    Unknown008 Posts: 8,076, Reputation: 723
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    #7

    Sep 23, 2009, 11:38 AM

    a) Nope. You need to take into consideration the directions as well. Ok, let's take the downwards direction as positive. The initial velocity is therefore -1.5m/s, the acceleration +10m/s^2 and the time is 2 s.

    b) Use the formula :

    Here, s is the displacement, from the point of release and the rest of the letters are the same as previously.

    Find the displacement 's'.

    Then, find the distance covered by the helicopter in 2 s during its ascension.

    Add the two distances to get the separation between the two.
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