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Math Help - motion of a projectile

  1. #1
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    motion of a projectile

    Here's a doozy for ya.
    A projectile is fired directly upward with a muzzle velocity of 860 feet per second from a height of 7 feet above the ground.

    A) Determine a function for the height of the projectile t seconds after it's released.
    B) How long does it take the projectile to reach a height of 100 feet on its way up?
    C) How long is the projectile in the air?



    So far, this is what I have for A.)
    X=(860cos90*) + and then I get lost. Any takers????
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by suzdisp View Post
    Here's a doozy for ya.
    A projectile is fired directly upward with a muzzle velocity of 860 feet per second from a height of 7 feet above the ground.

    A) Determine a function for the height of the projectile t seconds after it's released.
    B) How long does it take the projectile to reach a height of 100 feet on its way up?
    C) How long is the projectile in the air?



    So far, this is what I have for A.)
    X=(860cos90*) + and then I get lost. Any takers????
    The particle is fired directly upward, so we don't need to worry about x components or any angles. Thus the form for the height of the projectile is given by
    h = h_0 + v_0t + \frac{1}{2}at^2
    where h_0 is the height at t = 0, v_0 is the initial velocity, and a is the acceleration which in this case is -g = -32 ft/s^2.

    h = 7 + 860t - 16t^2

    See what you can do from there.

    -Dan
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  3. #3
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    Quote Originally Posted by suzdisp View Post
    Here's a doozy for ya.
    A projectile is fired directly upward with a muzzle velocity of 860 feet per second from a height of 7 feet above the ground.

    A) Determine a function for the height of the projectile t seconds after it's released.
    B) How long does it take the projectile to reach a height of 100 feet on its way up?
    C) How long is the projectile in the air?



    So far, this is what I have for A.)
    X=(860cos90*) + and then I get lost. Any takers????
    g=-\frac{32ft}{s^2} \mbox{ and } v_0=\frac{860ft}{2} \mbox{ and }x_0=7

    Using the kinematic equation x=\frac{1}{2}gt^2+v_0t+x_0

    gives us x=-16t^2+860t+7

    for b solve 100=-16t^2+860t+7

    for c solve 0=-16t^2+860t+7

    I hope this helps.
    Good luck.
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  4. #4
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    I'm sorry, I"m really bad at this math stuff. So for the equations you gave me for B) and C), do I just have to plug in a number for t to solve it?
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  5. #5
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    Quote Originally Posted by suzdisp View Post
    I'm sorry, I"m really bad at this math stuff. So for the equations you gave me for B) and C), do I just have to plug in a number for t to solve it?
    Solve them for t.

    Complete the square or use the quadratic formula.
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  6. #6
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    Quote Originally Posted by suzdisp View Post
    I'm sorry, I"m really bad at this math stuff. So for the equations you gave me for B) and C), do I just have to plug in a number for t to solve it?
    The equations given are quadratic equations. Have you been taught anything at all that's in this link?
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    I have had an overview of the quadratic equation, but I am just not getting it. It is hard because I am doing my school online, so there is really no tutor that I can sit down with and show me step by step what to do. I am one of those really dense people that needs to see EVERY step on how to solve the problems. I took my first test and failed it so now I'm trying to figure out how to do this WELL so I can pass my retake
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  8. #8
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    Quadratic formula

    if you have a quadratic equation in standard form
    ax^2+bx+c=0 Note: it must be equal to zero

    then

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

    Using this as an example if we have

    6x^2+7x-5=0

    The first thing we need to do is idendify a, b and c.

    a is the coeffient (the number in front of ) the x^2 term

    so a = 6

    b is the coeffient on the x term so b = 7

    c is the constant term -5 (note the sign comes with the number)

    so evaluating the quadratic formula we get..

    x=\frac{-(7) \pm \sqrt{(7)^2-4(6)(-5)}}{2(6)}

    Simplifying

    x=\frac{-7 \pm \sqrt{49+120}}{12}=\frac{-7 \pm \sqrt{169}}{12}

    x=\frac{-7 \pm 13}{12}

    so x= \frac{-7+13}{12}=\frac{1}{2} or

    x=\frac{-7-13}{12}=\frac{-20}{12}=-\frac{5}{3}

    Try this on the above problems.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by suzdisp View Post
    I have had an overview of the quadratic equation, but I am just not getting it. It is hard because I am doing my school online, so there is really no tutor that I can sit down with and show me step by step what to do. I am one of those really dense people that needs to see EVERY step on how to solve the problems. I took my first test and failed it so now I'm trying to figure out how to do this WELL so I can pass my retake
    Please don't take this the wrong way (I'm saying that a lot this morning!) but perhaps you should hire a personal tutor then? Just trying to think of what's best for your education.

    -Dan
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  10. #10
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    i cannot afford a personal tutor. why else would I be on here? duh just kidding!! anyways, okay. i tried to plug in my numbers for the problem and this is what I have so far:

    x=-860 +/- SR 860^2 - 4(16)(7)
    ----------------------
    2(16)
    x= -860 +/- SR 739600-448
    --------------
    2(16)

    x= -860 +/- SR 739152
    ----------
    2(16)

    x= -860 +/-860/32 ------->so 0
    --
    32



    or -1720
    ------ = -53.76 Am I moving in the right direction at all to solving this problem??
    32
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  11. #11
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    anybody...is that last reply right at all???
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  12. #12
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    Quote Originally Posted by topsquark View Post
    h = 7 + 860t - 16t^2
    -Dan
    Quote Originally Posted by suzdisp View Post
    x=-860 +/- SR 860^2 - 4(16)(7)
    ----------------------
    2(16)
    x= -860 +/- SR 739600-448
    --------------
    2(16)

    x= -860 +/- SR 739152
    ----------
    2(16)

    x= -860 +/-860/32 ------->so 0
    --
    32



    or -1720
    ------ = -53.76
    32
    Use more digits when calculating your answers.

    -16t^2 + 860t + 7 = 0

    So a = -16, b = 860, and c = 7.

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

    t = \frac{-860 \pm \sqrt{860^2 - 4(-16)(7)}}{2(-16)}

    t = \frac{-860 \pm \sqrt{739600 + 448}}{-32}

    t = \frac{-860 \pm \sqrt{740048}}{-32}

    t = \frac{-860 \pm 860.2604257}{-32}

    t = \frac{0.2604257}{-32} or t = \frac{-1720.2604257}{-32}

    t \approx -0.008138 or t \approx 53.7581

    -Dan
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