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,
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,
Using the quadratic formula,
To simplify by distributing the negative sign, it gets absorbed by the \(\pm\) symbol,
Then, it turns out that the positive solution corresponds to the positive sign \(+\text{.}\) In summary,
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,
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,
In summary,
Fact 1.8.1.
If an object is launched at (vertical) initial velocity \(v_0\) and initial height \(h_0\text{,}\) then the time it takes to hit the ground is,
Also, if \(v_0 > 0\text{,}\) then the object hits its maximum point at $t = \frac{v_0}{g}$, and its maximum height is,