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Math Help - Physics - Motion

  1. #1
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    Physics - Motion

    A man at the top of a building 20m high releases a stone from rest; 0.60 seconds later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. How long after the stone was dropped does the marble pass the stone?

    I know that after 0.6 seconds the stone is travelling at 6 ms-1, and because the acceleration is constant, this means that the marble is always travelling 2 ms-1 faster than the stone. I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.
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  2. #2
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    Re: Physics - Motion

    Quote Originally Posted by Fratricide View Post
    A man at the top of a building 20m high releases a stone from rest; 0.60 seconds later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. How long after the stone was dropped does the marble pass the stone?

    I know that after 0.6 seconds the stone is travelling at 6 ms-1, and because the acceleration is constant, this means that the marble is always travelling 2 ms-1 faster than the stone. I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.
    1st equation of motion

    \begin{align*}&t0=0 \\ &z1(0)= 20m \\&v1(0)=0m/s\\&a=-9.8m/s^2 \\&z1(t)=a(t-t0)^2+v1(0)(t-t0)+z1(0)=-9.8t^2+20\end{align*}

    2nd equation of motion

    \begin{align*}&t0=0.6 \\&z2(t0)= 20m \\&v2(t0)=-8m/s\\&a=-9.8m/s^2 \\&z2(t)=a(t-t0)^2+v(0)(t-t0)+z(0)=-9.8(t-0.6)^2-8(t-0.6)+20\end{align*}

    this should get you started.
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    Re: Physics - Motion

    Hello,
    Welcome to these forum sites, whatever you have asked, I don’t have
    right answer to this, but if you want to come India, you come
    to the,
    *************
    India Tour
    Thanks,
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    Re: Physics - Motion

    Quote Originally Posted by romsek View Post
    1st equation of motion

    \begin{align*}&t0=0 \\ &z1(0)= 20m \\&v1(0)=0m/s\\&a=-9.8m/s^2 \\&z1(t)=a(t-t0)^2+v1(0)(t-t0)+z1(0)=-9.8t^2+20\end{align*}

    2nd equation of motion

    \begin{align*}&t0=0.6 \\&z2(t0)= 20m \\&v2(t0)=-8m/s\\&a=-9.8m/s^2 \\&z2(t)=a(t-t0)^2+v(0)(t-t0)+z(0)=-9.8(t-0.6)^2-8(t-0.6)+20\end{align*}

    this should get you started.
    Thank you. I understand what to do with those equations, but could you explain to me in more depth the theory behind them and how you worked them out?
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  5. #5
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    Re: Physics - Motion

    Quote Originally Posted by Fratricide View Post
    Thank you. I understand what to do with those equations, but could you explain to me in more depth the theory behind them and how you worked them out?
    You can really find this derivation in a thousand places on the web worked out with pictures and animations etc. that I certainly can't show you here.

    Just google Newton's equations of motion.

    Incidentally I notice an error in the first equation

    z1(t)=a(t-t0)^2+v1(0)(t-t0)+z1(0)

    should read

    z1(t)=a(t-t0)^2+v1(0)(t-t0)+z1(t-t0)

    for the first equation they are the same thing since t0=0 but I should have been more careful. The same is true for the v1(0) there. It should be v1(t0).
    Last edited by romsek; January 3rd 2014 at 03:44 PM.
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  6. #6
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    Re: Physics - Motion

    Quote Originally Posted by Fratricide View Post
    A man at the top of a building 20m high releases a stone from rest; 0.60 seconds later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. How long after the stone was dropped does the marble pass the stone?

    I know that after 0.6 seconds the stone is travelling at 6 ms-1, and because the acceleration is constant, this means that the marble is always travelling 2 ms-1 faster than the stone.
    Are you rounding to one decimal place? Why would you do that? The downward acceleration, due to gravity, is 9.2 m/s2 so that after 0.6 seconds, the stone will be travelling at (9.2)(0.6)= 5.52 m/s downward. So the difference in speeds will be 8.00- 5.52= 2.48 m/s. In any case, the height of the stone, t seconds after it was thrown will be -4.6t^2+ 20 while the height of the marble will be -4.6(t- .6)^2- 8(t- .6)+ 20. The marble will "pass" the stone where their heights are equal: -4.6t^2+ 20= -4.6(t- .6)^2- 8(t- .6)+ 20. The t^2 terms will cancel out leaving a linear equation to solve for t.

    I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.
    Thanks from Fratricide
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  7. #7
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    Re: Physics - Motion

    All of my textbooks, assignments and exams for the course recognise the acceleration due to gravity as 10m/s2 (which obviously isn't accurate), and as such my physics teacher asks us to use that value in our calculations.

    As for the question, assuming gravity to be 10m/s2, I got:

    20 - 5t = 20 - 5(t - 0.6) - 8(t - 0.6)
    8t = 7.8
    t = 0.975 seconds

    EDIT: Just noticed that romsek used "a(t - t0)2​", whereas HallsofIvy used "(1/2)a(t - t0)2​".
    Last edited by Fratricide; January 5th 2014 at 03:07 PM.
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    Re: Physics - Motion

    Quote Originally Posted by Fratricide View Post
    All of my textbooks, assignments and exams for the course recognise the acceleration due to gravity as 10m/s2 (which obviously isn't accurate), and as such my physics teacher asks us to use that value in our calculations.

    As for the question, assuming gravity to be 10m/s2, I got:

    20 - 5t = 20 - 5(t - 0.6) - 8(t - 0.6)
    8t = 7.8
    t = 0.975 seconds

    EDIT: Just noticed that romsek used "a(t - t0)2​", whereas HallsofIvy used "(1/2)a(t - t0)2​".
    I suspect that was a typo by romsek. The distance traveled, from time [itex]t_0[/itex] to [itex]t[/itex], by an object with constant acceleration a is [itex]\frac{1}{2}a(t- t_0)^{1/2}[/tex].
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  9. #9
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    Re: Physics - Motion

    Let t=0 when marble is dropped.

    At t=0 the stone is at s0s with velocity v0s:
    s0s = (1/2)g(.6)2
    v0s = g(.6)

    Stone: ss = s0s + v0s(t) + (1/2)gt2
    Marble: sm = v0m(t) + (1/2)gt2

    ss=sm -> t = s0s/(v0m-v0s)

    Physically, this says, since marble has same acceleration as stone, marble will never catch up to stone unless its initial velocity is greater than the velocity of the stone at .6 secs.
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  10. #10
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    Re: Physics - Motion

    Something occurred to me and I missed “edit:”

    After finding t from last post, check that ss=(1/2)g(t+.6)2 < 20m, or else you will hit bottom before marble catches up with stone.
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