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    Different questions. Help pleaseAttachment 504683504685
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    (Original post by kiiten)
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    Different questions. Help pleaseAttachment 504683504685
    any workings?
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    For ques 3. I guessed s=3 and a=-9.8 i put that into s=ut+1/2at^2 and got 7.9=t (u+1/2t)

    For ques 1. I wasnt sure because i had the sane problem as before (too many unknowns)
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    (Original post by kiiten)
    For ques 3. I guessed s=3 and a=-9.8 i put that into s=ut+1/2at^2 and got 7.9=t (u+1/2t)

    For ques 1. I wasnt sure because i had the sane problem as before (too many unknowns)
    What can you say about the value of the J vector when it hits the ground, and how can we use this to find t
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    (Original post by samb1234)
    What can you say about the value of the J vector when it hits the ground, and how can we use this to find t
    Ah j becomes 0 so do you solve 6+9.5t-5t^2=0 to find t? If so could you x by -1 to make the 5 +ve?
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    (Original post by kiiten)
    Ah j becomes 0 so do you solve 6+9.5t-5t^2=0 to find t? If so could you x by -1 to make the 5 +ve?
    Yes
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    What about ques. 3??
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    Wait, how would you plot the path if you only have one value for t
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    Anyone? :,(
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    (Original post by kiiten)
    Anyone? :,(
    did you want help with q 1 or 3?
    i haven't done i and j notation yet but i have done suvat
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    (Original post by kiiten)
    For ques 3. I guessed s=3 and a=-9.8 i put that into s=ut+1/2at^2 and got 7.9=t (u+1/2t)

    For ques 1. I wasnt sure because i had the sane problem as before (too many unknowns)
    lol wat where did u get time from?
    I'll define up as positive

    If the max height the stone reaches is 5 m and the stone is thrown from 2 m above the ground the the distance it travels is 3m until it reaches its max height
    so s=3 which you got
    If up is positive and acceleration due to gravity acts down then
    a= -9.8
    u=u
    since the stone will stop momentarily for a split second at its max height
    v=0
    i'm sure you can work out the rest from there


    part b
    defining positive as down
    so on your diagram you'll be looking at the right hand side which has 5m labelled on the long line
    so finding the velocity of the stone when it hits the ground
    Remember now u=0 because we start by calculating from when the stone is momentarily still for a split second
    v=v
    a=9.8
    s=5
    i'm sure you can finish that one too

    part c
    try and do this one by yourself
    split the diagram into 2 where you split the diagram vertically at the point where the stone is momentarily still
    calculate the time for the first bit before the stone reaches max height and calculate time for after the stone has reached its max height and has reached the ground
    add them together and you should have your answer
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    (Original post by thefatone)
    did you want help with q 1 or 3?
    i haven't done i and j notation yet but i have done suvat
    Ok, ques 3. please - still need help on ques 1 though (for anyone else)
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    Zacken
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    (Original post by kiiten)
    Ok, ques 3. please - still need help on ques 1 though (for anyone else)
    Question 3 is just suvat, question 1 you just need to use the value of t you found in part a
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    (Original post by samb1234)
    Question 3 is just suvat, question 1 you just need to use the value of t you found in part a
    Im confused for part b on ques 3. Does s become 6? Also, for ques 1 you will only get 1 value so how do you plot a graph or are you supposed to do it when t is 1,2,3, etc.
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    No s would be 5. As far as plotting the path its just going to be a parabola, just plug some numbers in if you want a more accurate picture.
    (Original post by kiiten)
    Im confused for part b on ques 3. Does s become 6? Also, for ques 1 you will only get 1 value so how do you plot a graph or are you supposed to do it when t is 1,2,3, etc.
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    (Original post by samb1234)
    No s would be 5. As far as plotting the path its just going to be a parabola, just plug some numbers in if you want a more accurate picture.
    For the suvat im confused on part b. Could someone please explain. Do you use v^2 = u^2 + 2as and if you do s=5 and u=0 (5 is max. height) you get a -ve

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    (Original post by thefatone)
    Zacken
    Hey.

    (Original post by kiiten)
    For the suvat im confused on part b. Could someone please explain. Do you use v^2 = u^2 + 2as and if you do s=5 and u=0 (5 is max. height) you get a -ve

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    Plot the curve, just draw a parabola. There's no need for suvat in part b.

    Unless you're talking about "velocity of a stone as it hits the ground, in which case s=-2."
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    (Original post by Zacken)
    Hey.



    Plot the curve, just draw a parabola. There's no need for suvat in part b.

    Unless you're talking about "velocity of a stone as it hits the ground, in which case s=-2."
    So just use values of t like 1, 2 etc.

    Yes, it says velocity when the stone hits the ground. If you use suvat why would s=-2 dont you have to account for the distance from when its thrown up?
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    (Original post by kiiten)
    Yes, it says velocity when the stone hits the ground. If you use suvat why would s=-2 dont you have to account for the distance from when its thrown up?
    s is displacement not distance. If you got up x amounts and go down x+2 amounts then your displacement is -2.
 
 
 
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