Solution 7.4a - Revision history

Ian at 14:36, 14 May 2010

2010-05-14T14:36:54Z

←Older revision		Revision as of 14:36, 14 May 2010
Line 7:		Line 7:

	<math>\begin{align}		<math>\begin{align}
-	& ~~24+5t~~-{{t}^{\ 2}}=0 \\	+	& 8t-{{t}^{\ 2}}=0 \\
	& \\		& \\
-	& {{t}^{\ 2}}-~~5-24~~=0 \\	+	& {{t}^{\ 2}}-8t=0 \\
	\end{align}</math>		\end{align}</math>

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-	<math>~~\left(~~ t~~+3 \right)~~\left( t-8 \right)=0</math>	+	<math> t \left( t-8 \right)=0</math>


	giving solutions		giving solutions

-	<math>t=-3</math> and <math>t=8</math>	+	<math>t=0</math> and <math>t=8</math>

-	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.	+	The first solution is correct and shows that the frisby is thrown from ground level. We choose the second solution which gives the time when the frisby once more is at ground level.

	<math>t=8 \text { s}</math>		<math>t=8 \text { s}</math>

Ian: New page: The frisby hits the ground when the $\mathbf{k}$ term is zero. This gives $\begin{align} & 24+5t-{{t}^{\ 2}}=0 \\ & \\ & {{t}^{\ 2}}-5-24=0 \\ \end{align}$ ...

2010-04-11T16:15:21Z

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> ...

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>

This quadratic equation in
<math>t</math>
is to be solved. We factorise the equation.

<math>\left( t+3 \right)\left( t-8 \right)=0</math>

giving solutions

<math>t=-3</math> and <math>t=8</math>

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.

<math>t=8 \text { s}</math>

Solution 7.4a - Revision history

Ian at 14:36, 14 May 2010

Ian: New page: The frisby hits the ground when the \mathbf{k} term is zero. This gives \begin{align} & 24+5t-{{t}^{\ 2}}=0 \\ & \\ & {{t}^{\ 2}}-5-24=0 \\ \end{align} ...

Ian: New page: The frisby hits the ground when the $\mathbf{k}$ term is zero. This gives $\begin{align} & 24+5t-{{t}^{\ 2}}=0 \\ & \\ & {{t}^{\ 2}}-5-24=0 \\ \end{align}$ ...