Example  1


A soccer kick can be modelled by the function \(h\left( t \right) = -5t^2 + 30t\), where \(h\) is the height of the ball, in metres, and \(t\) is the time, in seconds, that the ball is in the air.

  1. How long will the ball be in the air?


    Step 1: Draw a diagram.

    Step 2: Write what you know.

    • A function is given: \(h\left( t \right) = -5t^2 + 30t\)
    • \(h\) is the height in metres
    • \(t\) is the time in seconds


    Step 3: Analyze how to solve the problem.


    Determine the amount of time the ball will be in the air.

    The ball will be at a height of \(0 \thinspace \rm{m}\) when it is kicked and when it lands. In between the time it is kicked and the time it lands, it will be in the air.

    Setting \(h(t) = 0\) will allow you to determine the time(s) the ball is at a height of \(0\).

    \(\begin{align}
     h\left( t \right) &= -5t^2 + 30t \\
     0 &= -5t^2 + 30t \\
     0 &= -5t\left( {t - 6} \right) \\
     \end{align}\)     


    \(\begin{array}{l}
     0 = -5t \\
     0 = t  \end{array}\)
    \(\begin{array}
     0 = t - 6 \\
     6 = t  \end{array}\)
    The two solutions to the equation are \(0\) and \(6\). The ball is kicked at \(0\) s and it lands at \(6\) s. As such, the ball will be in the air for \(6 – 0 = 6\) s.

  2. Determine the maximum height reached by the ball, to the nearest tenth of a metre.


    Step 1: Draw a diagram.

    Step 2: Write what you know.

    • A function is given: \(h\left( t \right) = -5t^2 + 30t\)
    • \(h\) is the height in metres
    • \(t\) is the time in seconds

    Step 3: Analyze how to solve the problem.

    Determine the maximum height reached by the ball. The maximum occurs at the vertex. The coordinates of the vertex can be determined if the equation of the axis of symmetry is known. The axis of symmetry is located at the midpoint between the \(x\)-intercepts.

    Step 4    : Solve the problem.

    The \(x\)-intercepts of the graph of the function are \(0\) and \(6\). Therefore, the axis of symmetry will be halfway between \(0\) and \(6\).

    \(t = \frac{{0 + 6}}{2} = 3\)

    Determine the coordinates of the vertex.

    \(\begin{align}
     h\left( t \right) &= -5t^2 + 30t \\
     h\left( 3 \right) &= -5\left( 3 \right)^2 + 30\left( 3 \right) \\
     h\left( 3 \right) &= -5\left( 9 \right) + 90 \\
     h\left( 3 \right) &= -45 + 90 \\
     h\left( 3 \right) &= 45 \end{align}\)

    The vertex is at \((3, 45)\). The \(y\)-coordinate of the vertex corresponds to the maximum height reached by the ball. The maximum height reached by the ball will be \(45 \thinspace \rm{m}\).
  3. When will the ball be at a height of \(12 \thinspace \rm{m}\)? Round to the nearest hundredth.

    The question is asking for the value of \(t\) when \(h(t) = 12\). Round to the nearest hundredth.

    \(\begin{align}
     h\left( t \right) &= -5t^2 + 30t \\
     12 &= -5t^2 + 30t \\
     0 &= -5t^2 + 30t - 12  \end{align}\)

    Use the quadratic formula to find the solution(s).

    \(a = -5, b = 30, c = -12\)

    \[\begin{align}
     x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
     x &= \frac{{ - 30 \pm \sqrt {30^2 - 4\left( { - 5} \right)\left( { - 12} \right)} }}{{2\left( { - 5} \right)}} \\
     x &= \frac{{ - 30 \pm \sqrt {900 - 240} }}{{ - 10}} \\
     x &= \frac{{ - 30 \pm \sqrt {660} }}{{ - 10}} \\
     x &\doteq 0.43, 5.57
     \end{align}\]


    The ball will be at a height of \(12 \thinspace \rm{m}\) at \(0.43\) s and at \(5.57\) s.
  4. State the restrictions, if any, on the domain and range for this scenario.

    The ball is kicked at \(t = 0\) and lands at \(t = 6\). The model is only useful when the ball is in the air, so the domain is \( \{t \thinspace | \thinspace 0 \le t \le 6, t \in \rm{R}\} \). The height of the ball ranges between \(0 \thinspace \rm{m}\) and \(45 \thinspace \rm{m}\), so the range is \(\{ (h(t)\thinspace | \thinspace 0 \le h(t) \le 45, h(t) \in \rm{R}\}\).