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Size:  521.7 KB Hi, for part c of this question, why can't you use the v^2=u^2+2as equation so that the previous answer is not used? I used this equation to work out v before the particle hits the ground but v is not the same as using v=u+at to work out v but why is that? Also the acceleration is worked out by using this equation
    but my equation is 0.6g-F= 0.6a since the weight acts downward and force opposes it. Why are they wrong? Thanks.
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    (Original post by coconut64)
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    What have you tried so far?
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    (Original post by SeanFM)
    What have you tried so far?
    I have just edited my post, please check it. Thanks.
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    (Original post by SeanFM)
    What have you tried so far?
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Size:  398.2 KB why can't I work out v using this equation ? Thanks.
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    (Original post by coconut64)
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Size:  398.2 KB why can't I work out v using this equation ? Thanks.
    You seem to be using the acceleration as 9.8, this is only true for freefall. It's sinking into the ground at some unknown acceleration that is affected by the resistive force.
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    (Original post by Zacken)
    You seem to be using the acceleration as 9.8, this is only true for freefall. It's sinking into the ground at some unknown acceleration that is affected by the resistive force.
    Hi again. This is to work out the velocity before it hits the ground.
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    (Original post by coconut64)
    Name:  1461525920576965410315.jpg
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Size:  521.7 KB Hi, for part c of this question, why can't you use the v^2=u^2+2as equation so that the previous answer is not used? I used this equation to work out v before the particle hits the ground but v is not the same as using v=u+at to work out v but why is that? Also the acceleration is worked out by using this equation
    but my equation is 0.6g-F= 0.6a since the weight acts downward and force opposes it. Why are they wrong? Thanks.
    If a is a negative value that may work (the acceleration will be acting in the same direction as the force, in other words it is deccelerating if you're looking at it from the direction from the weight).

    Perhaps you sued the wrong sign for a in v=u+at? (or inconsistent signs with v and u).
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    (Original post by SeanFM)
    If a is a negative value that may work (the acceleration will be acting in the same direction as the force, in other words it is deccelerating if you're looking at it from the direction from the weight).

    Perhaps you sued the wrong sign for a in v=u+at? (or inconsistent signs with v and u).
    I didn't use the v=u+at equation I used the other one but surely both can give me the same answer?
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    (Original post by coconut64)
    I didn't use the v=u+at equation I used the other one but surely both can give me the same answer?
    You used it to check, as far as I understand. And I was explaining why the two might not match up.
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    (Original post by SeanFM)
    You used it to check, as far as I understand. And I was explaining why the two might not match up.
    Which one would you use to check? The first equation I used was V^2=u^2+2as as I didn't want to use v in the equation and if I use the v=u+at equation, I get a different value for v though. I understand what you mean by the second part of that question.
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    (Original post by coconut64)
    Which one would you use to check? The first equation I used was V^2=u^2+2as as I didn't want to use v in the equation and if I use the v=u+at equation, I get a different value for v though. I understand what you mean by the second part of that question.
    I'm sorry, I can't get my head around what you're saying why are you using v=u+at, and what does the v represent in that equation? Are we talking about the times where it's all above ground, or the times where it's all below ground, or the time of both?
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    (Original post by coconut64)
    I didn't use the v=u+at equation I used the other one but surely both can give me the same answer?
    i remember using f=ma for this question :confused:
 
 
 
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