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    A particle is projected verically upwards with a speed of 30m/s from a point A. The point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4s. Calculate the value of h.

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    (Original post by snoopyx)
    A particle is projected verically upwards with a speed of 30m/s from a point A. The point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4s. Calculate the value of h.

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    At what's the maximum hight above A that the particle reaches? And at what time does it reach that hight?
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    I think this is simplest:

    First, consider the particle as it passes through B. How fast must it be going if it is to take 2.4 seconds before it comes back to B?
    Then use v^2 = u^2 + 2as to work out the velocity at A.
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    (Original post by DFranklin)
    I think this is simplest:

    First, consider the particle as it passes through B. How fast must it be going if it is to take 2.4 seconds before it comes back to B?
    Then use v^2 = u^2 + 2as to work out the velocity at A.
    Which equation would you use for the first part?
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    v=u+at, but i think you would use t as 1.2, not 2.4, because after 1.2 seconds of passing B, v=0
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    What would you use as a?
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    well the acceleration is gravity (pulling down and so it would be negative) so -9.81
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    9.8 is used
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    meh, i dunno, i'm used to doing these type of questions from AQA physics, and we always used 9.81, unless told otherwise. I think it depends on the exam board and stuff.
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    Yeah in physics we use 9.81 but as far as I know in maths 9.8 is used, for OCR anyway.
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    (Original post by snoopyx)
    A particle is projected verically upwards with a speed of 30m/s from a point A. The point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4s. Calculate the value of h.

    Thanks!
    v=u+at

    Let time to reach maximum height be  t then

    0=30-9.8t

    t=3.06122

    The first occasion that the particle is at height h is 3.05122-1.2=1.86122

    Sub this value into s=ut+\frac{1}{2}at^2
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    (Original post by steve2005)
    v=u+at

    Let time to reach maximum height be  t then

    0=30-9.8t

    t=3.06122

    The first occasion that the particle is at height h is 3.05122-1.2=1.86122

    Sub this value into s=ut+\frac{1}{2}at^2
    Hmm. a different approach - did you get 'h' as 38.81m?
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    (Original post by matty92)
    Hmm. a different approach - did you get 'h' as 38.81m?
    I got 38.8623 =38.86(2*d.p)
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    Ah very good! Yeh i quoted 'g' as 9.81, not 9.8, so yeh its the same
 
 
 
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