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Math Help - think its a quadratic equation but not sure hence why i need help

  1. #1
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    Question think its a quadratic equation but not sure hence why i need help

    A ball is thrown down at 72 km h1 speed from the top of a building. The
    building is 125 metres tall. The distance travelled before it reach the
    ground is as follows,

    s = u0t + 1/2gtsquared

    where:
    u0 = initial velocity (ms -1)
    g = acceleration due to gravity (10 ms -2)
    t = time (s)

    Find the time for the ball to drop to fifth of the height of the building.
    &
    Find the time for the ball to reach the ground.
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  2. #2
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    Re: think its a quadratic equation but not sure hence why i need help

    Quote Originally Posted by gordonmckee View Post
    A ball is thrown down at 72 km h1 speed from the top of a building. The
    building is 125 metres tall. The distance travelled before it reach the
    ground is as follows,

    s = u0t + 1/2gtsquared

    where:
    u0 = initial velocity (ms -1)
    g = acceleration due to gravity (10 ms -2)
    t = time (s)

    Find the time for the ball to drop to fifth of the height of the building.

    \color{red}{100 = 72t + 5t^2}

    &

    Find the time for the ball to reach the ground.

    \color{red}{125 = 72t + 5t^2}
    ... solve each equation for t
    Thanks from gordonmckee
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    Re: think its a quadratic equation but not sure hence why i need help

    need a little bit more help with that ...sorry
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    Re: think its a quadratic equation but not sure hence why i need help

    Quote Originally Posted by gordonmckee View Post
    A ball is thrown down at 72 km h1 speed from the top of a building. The
    building is 125 metres tall. The distance travelled before it reach the
    ground is as follows,

    s = u0t + 1/2gtsquared

    where:
    u0 = initial velocity (ms -1)
    g = acceleration due to gravity (10 ms -2)
    t = time (s)

    Find the time for the ball to drop to fifth of the height of the building.
    &
    Find the time for the ball to reach the ground.
    1. Convert u_0 = 72\ \frac{km}h = 20\ \frac ms

    2. \frac15 of the height is 25 m; thus the ball has travelled 100 m. Therefore you have to solve for t:

    100 = 20t + \frac12 \cdot 10\ \frac m{s^2} \cdot t^2

    3. When the ball hits the ground it has travelled 125 m; thus you have to solve for t:

    125 = 20t + \frac12 \cdot 10\ \frac m{s^2} \cdot t^2
    Thanks from gordonmckee
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    Re: think its a quadratic equation but not sure hence why i need help

    thanks but its the solving part im having trouble with
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    Re: think its a quadratic equation but not sure hence why i need help

    Quote Originally Posted by gordonmckee View Post
    thanks but its the solving part im having trouble with
    Use the quadratic formula:

    A quadratic equation


    ax^2+bx+c=0

    has the solution:

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

    Re-write your equation into the form above. Then plug in the values you know.

    Keep in mind that a negative time is not very plausible.
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    Re: think its a quadratic equation but not sure hence why i need help

    tried it and got 2.899 or -6.899.....am I doing this right or gone wrong somewhere?
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    Re: think its a quadratic equation but not sure hence why i need help

    That's not at all what I get. Could you show your calculations?

    (Of course, the quadratic equation will have two roots but the ball is not likely to hit the ground before it was thrown upward, is it?)
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    Re: think its a quadratic equation but not sure hence why i need help

    100 = 20t + 5tsquared
    a= 5
    b= 20
    c= -100

    x= -20 +or- the sqrt (20squared -4x5x-100) / 2 x 5
    = -20 + or - sqrt 2400 / 10
    =-20 + or - 48.99 / 10
    =2.899 or -6.899
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    Re: think its a quadratic equation but not sure hence why i need help

    Quote Originally Posted by HallsofIvy View Post
    That's not at all what I get. Could you show your calculations?
    Those do appear to be the (approximate) roots of the equation for the first problem, unless I'm missing something (maybe you didn't convert the initial velocity to m/s?). Naturally, we would discard the negative root.
    Thanks from gordonmckee
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