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Section 1.8 Applications of Quadratic Functions

Subsection 1.8.1 Application: Freefall Motion Revisited

The equation for the height of an object launched at initial velocity \(v_0\) and initial height \(h_0\text{,}\) is given by,

\begin{equation*} h(t) = -\frac{1}{2} t^2 + v_0 t + h_0 \end{equation*}

Then, consider when the object hits the ground, which corresponds to the \(t\)-intercept of the parabola (in particular, the \(t\)-intercept that is positive), i.e. the value of \(t\) such that \(h(t) = 0\text{.}\) That is, solve the equation,

\begin{equation*} -\frac{1}{2} t^2 + v_0 t + h_0 = 0 \end{equation*}

Using the quadratic formula,

\begin{align*} t \amp = \frac{-v_0 \pm \sqrt{(-v_0)^2 - 4\brac{-\frac{1}{2}g}(h_0)}}{2\brac{-\frac{1}{2}g}}\\ \amp = \frac{-v_0 \pm \sqrt{v_0^2 + 2gh_0}}{-g} \end{align*}

To simplify by distributing the negative sign, it gets absorbed by the \(\pm\) symbol,

\begin{equation*} t = \frac{v_0 \pm \sqrt{v_0^2 + 2gh_0}}{g} \end{equation*}

Then, it turns out that the positive solution corresponds to the positive sign \(+\text{.}\) In summary,

\begin{equation*} t = \frac{v_0 + \sqrt{v_0^2 + 2gh_0}}{g} \end{equation*}

Next, if the objects is launched upwards (so, \(v_0 > 0\)), then the object reaches a maximum point of motion before falling back down, and this point corresponds to the vertex of the parabola. The \(t\)-coordinate of the vertex is given by,

\begin{equation*} t = -\frac{b}{2a} = -\frac{v_0}{2\brac{-\frac{1}{2}g}} = \frac{v_0}{g} \end{equation*}

This is the time \(t\) where the object reaches its maximum height, if \(v_0 > 0\text{.}\) If \(v_0 \lt 0\text{,}\) the object never reaches a maximum height. Also, the maximum height is,

\begin{align*} h\brac{\frac{v_0}{g}} \amp = -\frac{1}{2} g \brac{\frac{v_0}{g}}^2 + v_0 \brac{\frac{v_0}{g}} + h_0\\ \amp = -\frac{1}{2} \cdot \frac{v_0^2}{g} + \frac{v_0^2}{g} + h_0\\ \amp = \frac{v_0^2}{2g} + h_0\\ \amp = \frac{v_0^2}{2g} + \frac{2gh_0}{2g}\\ \amp = \frac{v_0^2 + 2gh_0}{2g} \end{align*}

In summary,