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    Please look at the image attached

    Two small stones are projected horizontallyh towards each other at the same instant over horizontal ground. Initally the stones are 19.6m above the ground and are 50m apart. The intial speeds of the stones are 15 and v.

    What is the value of v if the stones collide at 4.9m above the ground?
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    What have you done so far? I got it, but I want to know where I should start.

    edit: Hello again :p:
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    (Original post by high_maintenance_girl)
    Please look at the image attached

    Two small stones are projected horizontallyh towards each other at the same instant over horizontal ground. Initally the stones are 19.6m above the ground and are 50m apart. The intial speeds of the stones are 15 and v.

    What is the value of v if the stones collide at 4.9m above the ground?
    using

    s = ut + \frac{1}{2}at^2

    you should know that vertical distance traveld is 19.6 - 4.9 = 14.7 and acceleration due to gravity is 9.8

    now you can find t.

    from this there are a few ways to work it out

    we can use

     s = v \times t

    we know v = 15 and we know t, so we can find the horizontal ditance travel by one stone, now we can use 50 - this answer, to find the ditance traveled by the other stone

    and using  speed = \frac{distance}{time}

    you get the answer (we've already figured out time.)
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    yes indeed hello there again!
    well tbh ive done nothing i tried a couple of things but now i just have papers with scribbles on it lol
    erm from a previous part of the question....ive correctly worked out...
    that for the left hand stone thingy
    the time between projection is 2 seconds
    and that the horizontal distance travelled is 30m
    that is all

    i think i should add you to my buddy list lol
    you are very helpful
    how old are you? and at what level are you studying maths?
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    thanks for the edit chaoslord i was very confused then lol
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    (Original post by high_maintenance_girl)
    thanks for the edit chaoslord i was very confused then lol
    so have you got it now?

    are the questions directed at me?
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    (Original post by Chaoslord)
    so have you got it now?

    are the questions directed at me?
    erm well i got time as root 3 seconds
    although im not familar with the s = v x t formula, i used it got the wrong answer

    and sorry no the questions were towards d-day - should've quoted
    but i was lazy and just did quick post
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    (Original post by high_maintenance_girl)
    erm well i got time as root 3 seconds
    although im not familar with the s = v x t formula, i used it got the wrong answer

    and sorry no the questions were towards d-day - should've quoted
    but i was lazy and just did quick post
    WOAHHHHHH take that back xD

    you ARE familiar with it! xD you probably dont understand the form

    it means, distance traveled = speed x time

    1 mph, traveling for one hour.

    you travel 1 mile xD

    the time between projection is 2 seconds
    and that the horizontal distance travelled is 30m
    that is all
    time = 2
    v = 15

    \therefore horizontal distance traveled = 30 which is what you got

    this wont work for the vertical component as there is an acceleration due to gravity
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    (Original post by Chaoslord)
    WOAHHHHHH take that back xD

    you ARE familiar with it! xD you probably dont understand the form

    it means, distance traveled = speed x time

    LMAO
    i am dumb :yep:
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    (Original post by high_maintenance_girl)
    LMAO
    i am dumb :yep:

    its okay you get it now though right? or would you like me to explain some more?
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    (Original post by Chaoslord)
    its okay you get it now though right? or would you like me to explain some more?
    explainy some more please?
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    (Original post by high_maintenance_girl)
    explainy some more please?
    Spoiler:
    Show

    considering the vertical motion
     s = ut + \frac{1}{2}at^2

     s = 19.6 - 4.9 = 14.7 (they were thrown from 19.6m and they meet at 4.9m so the distance traveled is 14.7m

    u = 0 since they fall from rest

    a = 9.8 this is the acceleration due to gravity

    subbing in we get


    14.7 =\frac{1}{2}(9.8)t^2

    3=t^2

    \therefore t = \sqrt3

    now consider the horizontal components

    for stone 1 (on the left)

    v = 15 and t = \sqrt3

    using distance traveled = speed \times time

    horizontal distance traveled by stone 1 = 15\sqrt3

    so the distance traveled by stone 2 = 50 - 15\sqrt3

    using speed = \frac{distance}{time}

    i get \frac{50 - 15\sqrt3}{\sqrt3}

    answer = about 13.86 ms-1


    sorry i kept confusing myself with typos xD
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    thank you very very much
    ive screenshotted that and kept it lol
    and what typos?
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    chaos, I'm glad you edited. Your original answer was nowhere near mine :p:
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    (Original post by high_maintenance_girl)
    thank you very very much
    ive screenshotted that and kept it lol
    and what typos?
    [qupte=D-Day]chaos, I'm glad you edited. Your original answer was nowhere near mine[/quote]

    roffle mao.

    accroding to my original calculations 3/2 time 2 = 3/4
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    :>.<:
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    yea just for that i deserved minus rep xD

    dday does your answer match mine?
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    I think so. I deleted it already :o:
 
 
 
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