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Thread: quick physics question

  1. #1
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    Default quick physics question

    yoh people!..just have a simple physics question and wanna clarify my doubts.

    A brick is dropped and takes 4 seconds to reach the ground . Assume that gravity = 10m/s

    when it reaches the ground. what speed would it be at.?

    what height was it thrown from?


    Thanks alot..

  2. #2
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    apply V = U + aT

    in this case.. initial speed U=0

    a is equal to g = 10 (like you said)

    and T is 4 seconds

    therefore V = (0) + (10)(4) = 40 m per second

  3. #3

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    height of 40meters?

    speed on ground is zero?

    okay someone beat me to the answer.. haha got show working summore..

  4. #4

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    Quote Originally Posted by slacker123
    apply V = U + aT

    in this case.. initial speed U=0

    a is equal to g = 10 (like you said)

    and T is 4 seconds

    therefore V = (0) + (10)(4) = 40 m per second

    V= Final velocity
    U= Initial velocity (in this case 0, cos it was just dropped, not thrown downwards)
    A= Acceleration (in this case 10m/s due to gravity)
    T= Time taken

    yepperz

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    haha..thanks guys..i also got the 40m/s..however whats the height it was dropped from..

    i was thinking..increase gradually...
    1, 10m/s
    2, 20m/s
    3, 30m/s
    4, 40m/s

    so add all that up..it would be dropped from 100m?

  6. #6

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    Taking downwards as positive

    Ignoring air-resistance,

    v = u + at
    v = 0 + (10)(4)
    v = 40m/s

    v = u + 2aS
    40 = 0 + 2(10)(s)
    1600 = 20s
    s = 80 m

    Final velocity at the moment before hitting the ground = 40 m s^-2
    Height = 80m
    Last edited by ipaquser; 22nd September 2004 at 09:48 PM.

  7. #7
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    Use v^2 = u^2 + 2as

    u^2 = 0 m/s

    a = 10 m/s^-2

    s = 0.5at^2 therefore 0.5x10x4^2 = 80m

    Subsituting s = 80m into equation v^2 = u^2 + 2as,
    v^2 = 2as = 2x10x80 = 1600
    v = sqrt 1600 = 40 m/s
    The equipment can only bring you so far - the rest of the photographic journey is done by you.

  8. #8
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    Quote Originally Posted by moos blues
    height of 40meters?

    speed on ground is zero?

    okay someone beat me to the answer.. haha got show working summore..

    This means that the object strikes the ground at the velocity of 40 meters per second.. the distance is computated using a different formula

    S = uT + 1/2(a)(T)^2 ( S equals ut plus half at squared)

    U = 0, T = 4, a = 10,

    therefore S = (0)(4) + (0.5)(10)(16) = 80

    distance.. or heigh dropped from the ground would then be 80 meters

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    Quote Originally Posted by SpitFir3
    haha..thanks guys..i also got the 40m/s..however whats the height it was dropped from..

    i was thinking..increase gradually...
    1, 10m/s
    2, 20m/s
    3, 30m/s
    4, 40m/s

    so add all that up..it would be dropped from 100m?
    To find height,

    use s=(at^2)/2

    a=10 m/s^2
    t= time taken to fall
    The equipment can only bring you so far - the rest of the photographic journey is done by you.

  10. #10

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    Actually, just apply any of these formulas.

    v = final velocity
    u = initial velocity
    a = acceleration
    s = distance
    t = time

    1. v = u + at
    2. v^2 = u^2 + 2aS
    3. s = ut + 0.5at^2

  11. #11

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    Quote Originally Posted by ipaquser
    Taking downwards as positive

    Ignoring air-resistance,

    v = u + at
    v = 0 + (10)(4)
    v = 40m/s

    v = u + 2aS
    40 = 0 + 2(10)(s)
    1600 = 20s
    s = 80 m

    Final velocity at the moment before hitting the ground = 40 m s^-2
    Height = 80m

    Good Ole....College days........Guess the laws of attraction between the Genders is not GMM/r^2 huh ??

    For what it's worth.....anyway.

  12. #12

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    Quote Originally Posted by microsmic
    Good Ole....College days........Guess the laws of attraction between the Genders is not GMM/r^2 huh ??

    For what it's worth.....anyway.
    Nope.. Its actually F = k(Q1Q2/r) =P

  13. #13

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    Quote Originally Posted by ipaquser
    Nope.. Its actually F = k(Q1Q2/r) =P
    Thanks for the correction......Einstein.

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    At the end of the day, most of the newtonian physics as we know it in classical mechanics is relatively accurate for small bodies of motion. Yet such Newtonian models of force, speed and distance is negated when it concerns very large and massive bodies. For example, planets, stars, galaxies and other universally large constants of motion. In such cases, we must consider the time lapse variable generated by the effect of gravity.

    As a real world example, most GPS satellites would give our locations wrongly by a few miles if they were not corrected by Einstein's General Theory of Relativity where the interaction of gravity, energy, mass, space and time affects each other simultaneously.

    Indeed, Classical Mechanics has been superseeded by Quantum Physics and Mechanics. For the newtonian model of F = GmM/r^2, it is invalid in the determinant of Mercury's orbit and revolution upon its axis due to the effects of time warp from the Sun. Thus we must be careful to note that what was taught to us during college days as well as first year undergraduate physics be taken with a pinch of salt at the higher levels.

    Then again, no doubt, the newtonian model is much simpler to understand, apply, use and to calculate from. The Einsteinian methologies exemplify extremely complex and complicated mathematics of at least graduate level stretching to post-Doctoral level.

    And at the end of the day. It is concluded that out of every student that took physics in secondary school and JC. 99.5% dislike the subject.

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    Quote Originally Posted by microsmic
    Good Ole....College days........Guess the laws of attraction between the Genders is not GMM/r^2 huh ??

    For what it's worth.....anyway.
    If GMm/r^2 were to work with the laws of attraction, we'd still be bachlors here!
    The equipment can only bring you so far - the rest of the photographic journey is done by you.

  16. #16
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    Newtonian Physics fails at the speed of light!

    Sighz... just when we thought we finished figuring out things...

    And light, depending on the frequency, has mass! Using E=mc^2, rearrange and fit E=hf, and you get m=hf/c^2!

    Ok I'm blabbering...
    The equipment can only bring you so far - the rest of the photographic journey is done by you.

  17. #17
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    Quote Originally Posted by nickmak
    If GMm/r^2 were to work with the laws of attraction, we'd still be bachlors here!
    In theory.. that would mean that a very heavy man and a very heavy women would have a greater chance of getting hitched given the same distance for a thin guy and a thin girl.

    Interesting thought, notable indeed.

    Perhaps at the end of the day, the theoretical law of attraction between the genders for the male and female would be

    male = {Pretty, Smart, Rich, Nice, Sweet, Cute, Submissive, etc}

    and perhaps:

    female = {Tall, Strong, Handsome, Rich, Smart}

    ideals of relationship often being just plain vagaries of methological differentiating in a "free economy of a big ocean of fish".

    Then again. perhaps only one fish out of a billion "suits the taste" of that one particular individual. Whenceforth money is earned by the wedding photographer

  18. #18

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    Quote Originally Posted by slacker123
    At the end of the day, most of the newtonian physics as we know it in classical mechanics is relatively accurate for small bodies of motion. Yet such Newtonian models of force, speed and distance is negated when it concerns very large and massive bodies. For example, planets, stars, galaxies and other universally large constants of motion. In such cases, we must consider the time lapse variable generated by the effect of gravity.

    As a real world example, most GPS satellites would give our locations wrongly by a few miles if they were not corrected by Einstein's General Theory of Relativity where the interaction of gravity, energy, mass, space and time affects each other simultaneously.

    Indeed, Classical Mechanics has been superseeded by Quantum Physics and Mechanics. For the newtonian model of F = GmM/r^2, it is invalid in the determinant of Mercury's orbit and revolution upon its axis due to the effects of time warp from the Sun. Thus we must be careful to note that what was taught to us during college days as well as first year undergraduate physics be taken with a pinch of salt at the higher levels.

    Then again, no doubt, the newtonian model is much simpler to understand, apply, use and to calculate from. The Einsteinian methologies exemplify extremely complex and complicated mathematics of at least graduate level stretching to post-Doctoral level.

    And at the end of the day. It is concluded that out of every student that took physics in secondary school and JC. 99.5% dislike the subject.

    Wow.....Great insight !!

    Anyhow, just got a feeling that the Theory of Relativity is in full swing here......is that 2 Einsteins that I'm seeing ??...from my point of view........Relatively ...Seriously....no flames intended.....only puns.

  19. #19

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    Quote Originally Posted by nickmak
    Newtonian Physics fails at the speed of light!

    Sighz... just when we thought we finished figuring out things...

    And light, depending on the frequency, has mass! Using E=mc^2, rearrange and fit E=hf, and you get m=hf/c^2!

    Ok I'm blabbering...
    you do that substitution when say a positron and electron annihilate each other and you're supposed to calculate the wavelength or frequency of the gamma ray emitted

  20. #20
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    Quote Originally Posted by nickmak
    Newtonian Physics fails at the speed of light!

    Sighz... just when we thought we finished figuring out things...

    And light, depending on the frequency, has mass! Using E=mc^2, rearrange and fit E=hf, and you get m=hf/c^2!

    Ok I'm blabbering...
    by the theorems of Max Planck

    E = hf + c

    C is a constant that is sometimes introduced to account for the Heisenberg Uncertainty Principle. At the same time, we must keep in mind that such an equation is very very simple. It is a fouding theorem for quantum mechanics that still works in some cases. At yet, most scientist, physicist, mathematicians and astronomers prefer to apply the Schrodeinger Equations to aid in accuracy in the calculation of minute sub atomic particles.

    Therefore, your conjecture should be along the line of

    E = (hf + c)/C^2

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