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    Hi,

    Has anybody done the June 2001 paper, particularly Question 4b/c
    If you haven't got it on you click the link below to access it.
    http://www.mathsexams.ukteachers.com...2_2001_Jun.pdf

    Answers to part A) (-25i - 5j)ms^1 B) 32.9m C) 51m


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    4b. As you need the vertical height, you only need consider the j component.
    At its greatest height, the ball's velocity will be 0ms^-1

    So 0=u^2-2as
    so s=u^2/2a
    =25^2/(2*9.8)
    =625/2g=31.9m (i hope you made a mistake on the answer you put)

    c. The time taken for ball to fall to 1m above ground is calculated as follows.

    s=ut+0.5at^2
    1=25t-4.9t^2
    so t=0.0403 or 5.0617s.
    The first time is when the ball passes 1m on the way up so we need the second value.

    Horizontally, the ball has travelled a distance:
    s=ut
    =10ms^-1 * 5.0617s
    =50.6m
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    (Original post by lgs98jonee)
    4b. As you need the vertical height, you only need consider the j component.
    At its greatest height, the ball's velocity will be 0ms^-1

    So 0=u^2-2as
    so s=u^2/2a
    =25^2/(2*9.8)
    =625/2g=31.9m (i hope you made a mistake on the answer you put)

    c. The time taken for ball to fall to 1m above ground is calculated as follows.

    s=ut+0.5at^2
    1=25t-4.9t^2
    so t=0.0403 or 5.0617s.
    The first time is when the ball passes 1m on the way up so we need the second value.

    Horizontally, the ball has travelled a distance:
    s=ut
    =10ms^-1 * 5.0617s
    =50.6m

    The mark scheme had 32.9m but it's only answers, not solutions
    It seems a little dodgy to me thou (the ms) cheers for parts b n c
 
 
 
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