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    Hey guys, so I'm a little stuck on a question from a past paper I'm doing as revision. I've attached a copy of the question, so hopefully you can get it alright. x

    Basically I've done part one with no trouble. And came out with the correct 5u/g.
    For the second part, I got the displacement as '30u^2/g' which is correct according to the mark scheme - so I'll skip over this part as well.

    It's the next section I'm having trouble with. "Find V, in terms of u." Sounds simple right? And I thought it would be, my working is as follows;

    Vertical Motion

    V = u + at
    V = 2.5u + g(5u/g)
    V = 2.5u + 5ug/g (here the 'g' can cancel out)
    V = 2.5u + 5u
    V = 7.5u or 7 1/2u

    That is what I got, but the answer in the mark scheme is actually 6.5u, not 7.5u. So can anyone explain to me why? What it is I have done wrong?

    The working out the mark scheme has is;

    Speed^2 = (6u)^2 + (2.5u)^2
    Speed = 6.5u

    But I can't work out what the formula they have used is?

    Thanks for reading, any help would be much appreciated.
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    (Original post by Pixie-Bee)
    Hey guys, so I'm a little stuck on a question from a past paper I'm doing as revision. I've attached a copy of the question, so hopefully you can get it alright. x

    Basically I've done part one with no trouble. And came out with the correct 5u/g.
    For the second part, I got the displacement as '30u^2/g' which is correct according to the mark scheme - so I'll skip over this part as well.

    It's the next section I'm having trouble with. "Find V, in terms of u." Sounds simple right? And I thought it would be, my working is as follows;

    Vertical Motion

    V = u + at
    V = 2.5u + g(5u/g)
    V = 2.5u + 5ug/g (here the 'g' can cancel out)
    V = 2.5u + 5u
    V = 7.5u or 7 1/2u

    That is what I got, but the answer in the mark scheme is actually 6.5u, not 7.5u. So can anyone explain to me why? What it is I have done wrong?

    The working out the mark scheme has is;

    Speed^2 = (6u)^2 + (2.5u)^2
    Speed = 6.5u

    But I can't work out what the formula they have used is?

    Thanks for reading, any help would be much appreciated.
    \cos^2 \alpha + \sin^2 \alpha = 1

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    I never thought of Trig Identities. xD But I don't understand how that shows the Final Velocity? o.O
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    (Original post by Pixie-Bee)
    I never thought of Trig Identities. xD But I don't understand how that shows the Final Velocity? o.O
    They give you 2 equations for V in the question - just square them and add them together
 
 
 
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