Solution 7.4a
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(New page: The frisby hits the ground when the <math>\mathbf{k}</math> term is zero. This gives <math>\begin{align} & 24+5t-{{t}^{\ 2}}=0 \\ & \\ & {{t}^{\ 2}}-5-24=0 \\ \end{align}</math> ...)
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Revision as of 16:15, 11 April 2010
The frisby hits the ground when the \displaystyle \mathbf{k} term is zero.
This gives
\displaystyle \begin{align}
& 24+5t-{{t}^{\ 2}}=0 \\
& \\
& {{t}^{\ 2}}-5-24=0 \\
\end{align}
This quadratic equation in
\displaystyle t
is to be solved. We factorise the equation.
\displaystyle \left( t+3 \right)\left( t-8 \right)=0
giving solutions
\displaystyle t=-3 and \displaystyle t=8
The first solution is mathematically correct, but is of no interest as we are looking for a positive time. We must choose the second solution.
\displaystyle t=8 \text { s}
