Putting it Together: Quadratic Equations and Complex Numbers

[latex]h(t)=-10\frac{m}{ s^2}t^2+v_{o}t+h_{o}[/latex]

She had labeled the equation with arrows in the following way: [latex]-10\frac{m}{ s^2}[/latex] is gravity, [latex]v_{o}[/latex] is the initial speed of the object, [latex]h_{o}[/latex] is the initial height of the object.

Joan and Hazel decide to define some of the unknowns in the equation. Since they started from a standing position, they decide that [latex]v_{o}=0[/latex].  Then, they estimate that the hight of the cliff was about [latex]15[/latex] meters (or about 45 feet), so [latex]h_{o}=15[/latex].  They use substitution to replace these values in the equation and simplify:

[latex]\begin{array}{c}h(t)=-10\frac{m}{ s^2}{ t^2}+0\cdot{0}\cdot{t}t+15m\\h(t)=-10\frac{m}{ s^2}{ t^2}+15m\end{array}[/latex]

Removing the units: [latex]h(t)=-10{t^2}+15[/latex]

"Now what?" asks Hazel.  "I don't remember what we did from here." Joan takes over because she has just finished a module on quadratic equations in her math class, and she knows this is a quadratic equation.

"Our original question was, how long did it take us to fall from the cliff to the water, right? So that means that our height above ground at that time would be equal to zero.  I think we should set the equation equal to zero and solve for t."

[latex]h(t)=-10t^{2}+15=0[/latex]

[latex]-10{t}^{2}=-15[/latex]

[latex]{t}^{2}=\frac{3}{2}[/latex]

[latex]\sqrt{ t^2}=\sqrt{\frac{3}{2}}[/latex]

Using a calculator, they evaluate [latex]\sqrt{\frac{3}{2}}[/latex] and get [latex]1.22[/latex].

"Wow! When you are falling it seems longer than [latex]1.22[/latex] seconds!" says Hazel.

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