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    I dont understand the second part to iii) can somebody help please will be so much appreciated
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    The parts are travelling horizontally, maybe at different speeds, but they are travelling horizontally, not vertically. They both start at the same vertical point, so assuming in this model that the vertical change is nothing, they both remain at the same height.
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    (Original post by BlueHighlighter)
    The parts are travelling horizontally, maybe at different speeds, but they are travelling horizontally, not vertically. They both start at the same vertical point, so assuming in this model that the vertical change is nothing, they both remain at the same height.
    I get that but don't understand the bit just after it - the hence bit can you help me there please?
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    s=ut+1/2at^2 and substitute the values in
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    (Original post by Razz13)
    s=ut+1/2at^2 and substitute the values in
    Can you pls explain a little bit more apart from the formula
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    use s=1/2(v+u)*t for part (i) and then i think u use s=ut+1/2*at^2 for part (ii)
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    (Original post by Rauhan)
    use s=1/2(v+u)*t for part (i) and then i think u use s=ut+1/2*at^2 for part (ii)
    part iii please not those -the hence bit
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    yea sorry i didnt read the line you wrote at the end of the picture; lemme have a look at it
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    ok so since you know these things abt the parts

    part 1
    s=?
    u=21
    a= 0 since its projecting horizontally
    t=t

    part 2
    s=?
    u=5.25m/s
    a= 0 since its projecting horizontally
    t=t

    now use the equation s=ut+1/2at^2 to find the distance in terms of t and u will end up having 2 simultanious equations
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    i am still confused abt the part where they ask for the reason
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    (Original post by Rauhan)
    ok so since you know these things abt the parts

    part 1
    s=?
    u=21
    a= 0 since its projecting horizontally
    t=t

    part 2
    s=?
    u=5.25m/s
    a= 0 since its projecting horizontally
    t=t

    now use the equation s=ut+1/2at^2 to find the distance in terms of t and u will end up having 2 simultanious equations
    Hi! Thanks for replying. Just attached ms below
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    Theyve done this weird thing for their method
    0.75 x 21t =15.75t -->this doesn't make much sense
    Do you know what they've done here?
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    dosent make sense to me either cu in part (iii) they never mentioned anything abt the distance except for finding it in terms of t and then in part (iv) finding the distance through simultaneous equations
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    what year is this question from
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    (Original post by Rauhan)
    what year is this question from
    its 2008 and yhh its confusing what they did there in the ms
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    i just figured it out; so whats the value of t u got in (i)
 
 
 

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