From Mechanics

Revision as of 16:15, 11 April 2010 (edit) Ian (Talk | contribs) (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> ...) ← Previous diff		Current revision (14:36, 14 May 2010) (edit) (undo) Ian (Talk | contribs)
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>
Line 18:		Line 18:
-	<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>

---

Current revision

The frisby hits the ground when the \displaystyle \mathbf{k} term is zero.

This gives

\displaystyle \begin{align} & 8t-{{t}^{\ 2}}=0 \\ & \\ & {{t}^{\ 2}}-8t=0 \\ \end{align}

This quadratic equation in \displaystyle t is to be solved. We factorise the equation.

\displaystyle t \left( t-8 \right)=0

giving solutions

\displaystyle t=0 and \displaystyle t=8

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.

\displaystyle t=8 \text { s}