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Mechanics - M1: Motion Under Gravity. watch

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    attemped most the question, just need some reassurance.

    "A ball is thrown vertically upards at 4.9ms^{-1} from a window which is 5m above the horizontal ground. By modelling the ball as a particle moving freely under gravity, find:

    a, the greatest height above the ground attained by the ball.
    b, the time taken by the ball to reach the ground.
    c, state two assumptions made.

    u=4.9
    a=-9.8
    v=0

    v^{2}=u^{2}+2as

    s=\frac{-24.01}{-19.6}

    s=1.225

    greatest height above ground = 1.225 + 5 = 6.225m

    b,

    v=u+at

    0=4.9+9.8t

    t=\frac{-4.9}{9.8}

    t=-0.5

    c,i, there is no air resistance,
    ii, gravity is constantly 9.8 m/s/s
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    a) is correct

    b) is wrong - the velocity is not zero when it hits the ground. Use s = -5 m to represent ground level.

    c) i) is correct

    c) ii) should be the ball is modelled as a particle

    Don't write m/s/s - use m/s^2 or ms^{-2}
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    (Original post by Mr M)
    a) is correct

    b) is wrong - the velocity is not zero when it hits the ground. Use s = -5 m to represent ground level.

    c) i) is correct

    c) ii) should be the ball is modelled as a particle

    Don't write m/s/s - use m/s^2 or ms^{-2}
    hmm...

    v^{2}=(4.9)^{2}+2(-9.8)(-5)

    v^{2}=122.01

    v=11.04581

    v=11.0 (3sf)

    ????
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    (Original post by KeineHeldenMehr)
    hmm...

    v^{2}=(4.9)^{2}+2(-9.8)(-5)

    v^{2}=122.01

    v=11.04581

    v=11.0 (3sf)

    ????
    There are two square roots. You want the negative one to represent the direction of the velocity downwards. But the magnitude is correct.
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    (Original post by Mr M)
    There are two square roots. You want the negative one to represent the direction of the velocity downwards. But the magnitude is correct.
    v^{2}=122.01

    v=\pm 11.04581

    so
    v=11.0
    v=-11.0

    might have got the wrong end of the stick...

    i just realised i still have to work out T..

    ok,
    v=u+at?
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    No that's correct.

    You just want to state the negative velocity.

    The positive velocity does not actually occur. It is produced by the symmetry of the mathematics and would occur at a negative time (I mean it would happen at a time before the ball is even thrown upwards!!!).
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    I may be wrong here, but Mr M, are you sure you should you say "assume the ball is modelled as a particle" if that's what they've asked you to do in the question? It's been a while since M1, so I may be hazy.
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    (Original post by KeineHeldenMehr)
    v^{2}=122.01

    v=\pm 11.04581

    so
    v=11.0
    v=-11.0

    might have got the wrong end of the stick...

    i just realised i still have to work out T..

    ok,
    v=u+at?
    Yes, that is fine.

    You could have picked a different SUVAT equation that would have given you the time immediately. Which one?
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    (Original post by rt6)
    I may be wrong here, but Mr M, are you sure you should you say "assume the ball is modelled as a particle" if that's what they've asked you to do in the question? It's been a while since M1, so I may be hazy.
    No you are right. I didn't read the question. The ones I stated are the two standard assumptions. In this case, the OP's answers were fine.
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    (Original post by Mr M)
    No that's correct.

    You just want to state the negative velocity.

    The positive velocity does not actually occur. It is produced by the symmetry of the mathematics and would occur at a negative time (I mean it would happen at a time before the ball is even thrown upwards!!!).
    ok,

    v=-11
    u=4.9
    a=-9.8

    v=u+at

    -11=4.9+(-9.8)t

    t= \frac{-15.9}{-9.8}

    t=1.62 (3sf)
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    (Original post by Mr M)
    Yes, that is fine.

    You could have picked a different SUVAT equation that would have given you the time immediately. Which one?
    s=(\frac{u+v}{2})t?
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    Yes, it is actually 1.63 seconds (3 sf) - try to avoid using rounded numbers in calculations as it introduces rounding errors.
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    (Original post by KeineHeldenMehr)
    s=(\frac{u+v}{2})t?
    The information you knew was:

    u=4.9

    a=-9.8

    s=-5

    t=?

    So you wanted to pick s=ut+0.5at^2

    This did not require you to find v first.
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    (Original post by Mr M)
    Yes, it is actually 1.63 seconds (3 sf) - try to avoid using rounded numbers in calculations as it introduces rounding errors.
    what did you substitute v as?
    how many significant figures did you enter v to?
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    (Original post by Mr M)
    The information you knew was:

    u=4.9

    a=-9.8

    s=-5

    t=?

    So you wanted to pick s=ut+0.5at^2

    This did not require you to find v first.
    ah, for future purposes, ill look out more to use it!
    thanks.
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    (Original post by KeineHeldenMehr)
    what did you substitute v as?
    how many significant figures did you enter v to?
    Use as many as you can (all the numbers on your calculator if possible).

    Better still, use the surd form for v as it is exact.

    v=\frac{-7\sqrt{249}}{10}

    Even better than that would be to use s=ut+0.5at^2 which avoids the need to find v or introduce any rounding errors at all.
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    (Original post by KeineHeldenMehr)
    ah, for future purposes, ill look out more to use it!
    thanks.
    you are welcome
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    puzzled if i should make distance negative of not..

    "a competitor makes a dive from a high springboard into a diving pool. She leaves the springboard vertically with a speed of 4ms^{-1} upwards. When she leaves, she is 5m above the surface of the pool. the diver is modelled as a particle moving veritcally under gravity alone and it is assumed that she does not hit the springboard as she descends. Find:

    1, her speed when she reaches the surface of the pool,
    2, the time taken to reach the surface of the pool,
    3, state two physical factors that have been ignored in the model.

    ok, i've figured out that

    s=-5
    u=4
    v=?
    a=9.8
    t=?

    just need some checking here and there.

    1,
    v^{2}=u^{2}+2as

    v^{2}=4^{2}+2(9.8)(5)

    is it suppose to be -5?
    not sure, had that feeling since it's above.

    v^{2}=114

    v=10.7 (3sf)

    2,

    s=ut+\frac{1}{2}at^{2}

    5=(4)(t)+\frac{1}{2}(9.8)(t)^{2}

    10=8t+9.8t^{2}

    9.8^{2}+8t-10

    quadratic formuale......

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    x = \frac{-8 \pm \sqrt{8^2-4(9.8)(10)}}{2(9.8)}

    x = \frac{-8 \pm \sqrt{64+392}}{19.6}

    x = \frac{-8 \pm 21.4}{19.6}

    two answers:

    0.68s, -1.5s

    therefore, 0.68 , since -ve a second is impossible?

    3, no air resistance...

    not sure about a 2nd point.
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    Remember to keep your signs correct. If you're taking up as positive, u is positive, and g and s are negative.
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    (Original post by Scipio90)
    Remember to keep your signs correct. If you're taking up as positive, u is positive, and g and s are negative.
    so 1,2 &3 are wrong... following this statement..
    actually 1 stays the same since:
    v=\sqrt {4^{2}+2(-9.8)(-5)}

    the minuses cause them to add.

    2,

    -5=4t-\frac{-9.8t^{2}}{2}

    heading the right direction?

    for 3, any ideas on a 2nd point i could include?
 
 
 
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