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Math Help - Gravity motion mechanics 1

  1. #1
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    Gravity motion mechanics 1

    Could do with some help

    In a volcanic eruptioni a rock is ejected vertically from the crater and rises to a height of 25 kilometres. Find the speed with which it emerges from the crater, and time it takes to reach this height.

    Would appreciate any help

    thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by gracey View Post
    Could do with some help

    In a volcanic eruptioni a rock is ejected vertically from the crater and rises to a height of 25 kilometres. Find the speed with which it emerges from the crater, and time it takes to reach this height.

    Would appreciate any help

    thanks
    There are four basic kinematics equations for motion under constant acceleration. You are going to use two of them here.

    For the first part we know that
    v^2 = v_0^2 + 2a(y - y_0)
    So set the zero level at the top of the volcano and upward as positive. Notice that at the top of the motion the speed is 0 m/s. So we get
    0 = v_0^2 - 2gy
    which you can solve for v_0.

    For the second part we also know that
    v = v_0 + at
    and again v = 0 m/s so
    0 = v_0 - gt
    You know v_0 from the first part, so now just solve for t.

    -Dan
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  3. #3
    MHF Contributor arbolis's Avatar
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    Just for fun, as g (the gravitational force on the Earth's surface) is not constant over these 25 \text{km}, you can also find v_0 with the following formula : E=\frac{1}{2}mv^2-\frac{GmM_E}{r}. Considering the Earth as a sphere with radius =6400 \text{km}, G=6.67\times10^{-11}\text{N}\frac{m^2}{kg^2} and M_E=5.97\times 10^{24}\text{kg} I found that v_0 is worth 695.84 \frac{m}{s}. Compare this result with yours, I think mine should be a bit lesser than yours.
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  4. #4
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    Hello, gracey!

    In a volcanic eruption, a rock is ejected vertically from the crater
    and rises to a height of 25 kilometres.

    Find the speed with which it emerges from the crater,
    and time it takes to reach this height.

    The equation is: . h(t) \;=\;h_o + v_ot - 4.9t^2

    where: . \begin{Bmatrix} h &=& \text{height of rock} \\ h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix} . measured in meters and seconds


    Assume h_o = 0
    . . We will measure the rock's height from the top of the volcano.

    The equation is: . h(t) \;=\;v_ot - 4.9t^2

    This is a down-opening parabola; its peak is at its vertex.
    . . v \:=\:\frac{\text{-}b}{2a}

    With a = \text{-}4.9 and b = v_o, we have: . t \:=\:\frac{\text{-}v_o}{2(\text{-}4.9)} \:=\:\frac{v_o}{9.8} seconds. .[1]

    The maximum height is: . h\left(\frac{v_o}{9.8}\right) \;=\;v_o\left(\frac{v_o}{9.8}\right) - 4.9\left(\frac{v_o}{9.8}\right)^2 \;=\;\frac{v_o}{19.6} meters


    Since the maximum height is 25 km = 25,000 m,
    . . \frac{v_o}{19.6} \:=\:25.000 \quad\Rightarrow\quad v_o \:=\:490,000\text{ m/sec}


    Therefore, the initial velocity is: . \boxed{490\text{ km/sec}}


    Substitute into [1]: . t \;=\;\frac{490,000}{9.8} \;=\;\boxed{50,000\text{ seconds}}

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  5. #5
    MHF Contributor arbolis's Avatar
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    What?! Soroban, an initial speed of almost 500 \text{ km/s}! And it takes 50 sec to reach only 25 \text{ km}! You certainly mean 490 \frac{m}{s} which surprises me because I thought it would have been more than 496\frac{m}{s}.
    I don't see where is my error...
    Last edited by arbolis; September 12th 2008 at 11:44 AM.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post

    The maximum height is: . h\left(\frac{v_o}{9.8}\right) \;=\;v_o\left(\frac{v_o}{9.8}\right) - 4.9\left(\frac{v_o}{9.8}\right)^2 \;=\;\frac{v_o}{19.6} meters


    Since the maximum height is 25 km = 25,000 m,
    . . \frac{v_o}{19.6} \:=\:25.000 \quad\Rightarrow\quad v_o \:=\:490,000\text{ m/sec}


    Therefore, the initial velocity is: . \boxed{490\text{ km/sec}}
    There was a typo or something here.


    Substitute into [1]: . t \;=\;\frac{490,000}{9.8} \;=\;\boxed{50,000\text{ seconds}}

    h\left(\frac{v_o}{9.8}\right) \;=\;v_o\left(\frac{v_o}{9.8}\right) - 4.9\left(\frac{v_o}{9.8}\right)^2 \;=\; \frac{v_0^2}{19.6}

    So
    \frac{v_0^2}{19.6} = 25000 \implies v_0 = 700~m/s

    -Dan
    Last edited by topsquark; September 12th 2008 at 12:16 PM.
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  7. #7
    MHF Contributor arbolis's Avatar
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    topsquark, I think you are wrong with Soroban, I graphed the functions of position versus time and my answer was closer to 25000 m while 522 m/s was closer to 13000 m. Can you check that out once again?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by arbolis View Post
    topsquark, I think you are wrong with Soroban, I graphed the functions of position versus time and my answer was closer to 25000 m while 522 m/s was closer to 13000 m. Can you check that out once again?
    You are correct. I must have punched the numbers in wrong. I have fixed it in my response.

    The point is that Soroban forgot to square the v_0 in his height equation. And notice that 490 km/s is significantly higher than escape velocity, so it won't have a max. height.

    -Dan
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