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Math Help - Mechanics question

  1. #1
    Member
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    Sep 2008
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    Mechanics question

    Hi could anyone help with this? I can do part A but not part B.

    A pebble is thrown vertically upwards from a point A with initial velocity 11.2 m/s

    A) Find the greatest point above A that can be reached by pebble B
    (I make it 6.4m)

    The point B is 9m vertically above point A. At the same time that P was thrown from A, another pebble Q was thrown vertically downwards from B with initial velocity 0.8 m/s. The two pebbles collide after t seconds at a height h metres above point A.

    B) Calculate the values of t and h

    Not sure how to even start this bit :/

    thanks
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  2. #2
    MHF Contributor
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    Umm, this projectile thing is also Mechanics? I remembered it as a Physics thing.
    Umm, it is about vectors. So, yes, it must be Mechanics also.

    Yes, you are right that the pebble reaches a maximum of 6.4m in part A.

    For the part B, it's almost the same as in part A. It is vector thing.
    As with vectors, the directions of the flights are very important.

    s = Vo*t +(1/2)a*t^2
    That's the basic formula for distance or displacement.

    In the going vertically up, the acceleration is g and it is negative because the going up is measured as positive.
    In the going vertically down, the acceleration is g also....but now it is positive because the going down is measured as positive here.

    g is always going down, as you know.

    For the going up,
    h = 11.2*t -(1/2)(9.8)t^2 -----------(1)

    For the going down,
    h1 = 0.8*t +(1/2)(9.8)t^2 ----------(2)

    h +h1 = 9 m
    So, adding (1) and (2)
    h +h1 = 12t
    t = 9/12 = 0.75 sec ---------answer.

    And so,
    h = 11.2(0.75) -(1/2)(9.8)(0.75)^2
    h = 5.644 m above point A ----------answer.
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  3. #3
    Member
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    Sep 2008
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    yeah physics and mechanics are very closely related (both use SUVAT equations) thanks a lot for the reply, seems my posts rarely get replied to recently lol so yeah thanks a lot
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