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    If a particle goes vertically up from ground (A) and it arrives to highest point (B). Then comes back from B to A.

    Is the time for the ball going up from A to B equal to the time going down from B to A?
    This looks wrong because for A to B gravity acts against it and for B to A gravity acts with it.
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    Yes it is. The acceleration (due to gravity) is always acting downwards on the particle; when it's travelling up it has an upwards velocity. It doesn't sound right, I know what you mean, but if you think about it, it travels upwards from A, then slows down which means it must have a force acting against it (gravity).
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    its true since its symmetrical

    do a question on a practice paper

    are you doing the exam on 23rd jan - edexcel

    ryan
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    (Original post by ryanb97)
    its true since its symmetrical

    do a question on a practice paper

    are you doing the exam on 23rd jan - edexcel

    ryan
    Yes im doing the exam on 23rd jan - edexcel.
    Can you give me a question which is similar so I can do it. :cool:
    which practice paper?
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    A good way to see this in action is to set up an imagined situation yourself where you could see whether this is the case or not. For instance, imagine a ball being fired straight up from the ground with u=40m/s. Now, a=-9.8, v=0. Use the equations of motion to work out the time taken for the ball to come to rest (instantaneously), then work out how long it takes to hit the ground again. The answer will be the same (I got 4.08s).

    Here's another way to look at it: s=-\frac{1}{2}at^2 for the ball going up, and s=\frac{1}{2}at^2 for the ball coming down. And s and a have the same values for both the ball coming up, and the ball going down(except a is negative in one, and positive in the other). So they'll both give the same answer for t.
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    (Original post by JuliusDS92)
    A good way to see this in action is to set up an imagined situation yourself where you could see whether this is the case or not. For instance, imagine a ball being fired straight up from the ground with u=40m/s. Now, a=-9.8, v=0. Use the equations of motion to work out the time taken for the ball to come to rest (instantaneously), then work out how long it takes to hit the ground again. The answer will be the same (I got 4.08s).

    Here's another way to look at it: s=-\frac{1}{2}at^2 for the ball going up, and s=\frac{1}{2}at^2 for the ball coming down. And s and a have the same values for both the ball coming up, and the ball going down(except a is negative in one, and positive in the other). So they'll both give the same answer for t.
    Let me try in two methods:
    Method 1:
    u= 40m/s, a=-9.8 and v=0
    therefore t=(v-u)/a=(0-40)/-9.8= 4.08s
    From A to A its 4.08x2 = 8.16s

    Method 2:
    u = 40m/s a=-9.8 and s=0
    s=ut+0.5at^2
    0= 40t+0.5x-9.8xt^2 => t(40+0.5x-9.8xt)=0
    t=0s or t=-40/(-9.8x0.5)=8.16s
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    (Original post by Namod)
    Let me try in two methods:
    Method 1:
    u= 40m/s, a=-9.8 and v=0
    therefore t=(v-u)/a=(0-40)/-9.8= 4.08s
    From A to A its 4.08x2 = 8.16s

    Method 2:
    u = 40m/s a=-9.8 and s=0
    s=ut+0.5at^2
    0= 40t+0.5x-9.8xt^2 => t(40+0.5x-9.8xt)=0
    t=0s or t=-40/(-9.8x0.5)=8.16s
    That's not what I did, but it looks right. What's important is if by doing this you can really get a sense of why it's right, rather than knowing it's right "just because you've been told so".
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    (Original post by JuliusDS92)
    A good way to see this in action is to set up an imagined situation yourself where you could see whether this is the case or not. For instance, imagine a ball being fired straight up from the ground with u=40m/s. Now, a=-9.8, v=0. Use the equations of motion to work out the time taken for the ball to come to rest (instantaneously), then work out how long it takes to hit the ground again. The answer will be the same (I got 4.08s).

    Here's another way to look at it: s=-\frac{1}{2}at^2 for the ball going up, and s=\frac{1}{2}at^2 for the ball coming down. And s and a have the same values for both the ball coming up, and the ball going down(except a is negative in one, and positive in the other). So they'll both give the same answer for t.
    How can you do that?
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    (Original post by Namod)
    Yes im doing the exam on 23rd jan - edexcel.
    Can you give me a question which is similar so I can do it. :cool:
    which practice paper?
    go through the last 3 years of past papers on www.examsolutions.net

    its the best way to revise!! well for me

    look at his videos if you get a question wrong..it will explain everything you need for the exam.. and for an A

    oh and do the june 2009 paper...crucial!!... use the guys videos .. i mean seriously step by step explanations of everything!

    Ryan
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