Solution 7.4c - Revision history

Ian at 14:56, 14 May 2010

2010-05-14T14:56:21Z

←Older revision		Revision as of 14:56, 14 May 2010
Line 1:		Line 1:
	We insert		We insert
	<math>t=0\ \text{s}</math>		<math>t=0\ \text{s}</math>
-	in the expression for the position vector to find the point where the frisby ~~waqs~~ thrown from.	+	in the expression for the position vector to find the point where the frisby was thrown from.


Line 7:		Line 7:
	& \mathbf{r}=\left( 5\times 0-{{0}^{2}} \right)\mathbf{i}+\left( 7\times 0 \right)\mathbf{j}+0\mathbf{k} \\		& \mathbf{r}=\left( 5\times 0-{{0}^{2}} \right)\mathbf{i}+\left( 7\times 0 \right)\mathbf{j}+0\mathbf{k} \\
	& \\		& \\
-	& =~~-24~~\mathbf{i}+56\mathbf{j}+0\mathbf{k}\ \text{m} \\	+	& =0\mathbf{i}+0\mathbf{j}+0\mathbf{k}\ \text{m}\\
	\end{align}</math>		\end{align}</math>

-	~~The~~ point where the frisby was thrown is the origin. ~~This is obtained by putting~~	+	Thus the point where the frisby was thrown from is the origin.
-	~~<math>t=0</math>~~	+
-	in the ~~expression for~~	+	Using the result from part b)
-	<math>\~~mathbf~~{r}~~</math>~~	+
-	~~giving <math>~~\mathbf{r} =0\mathbf{i}+0\mathbf{j}+0\mathbf{k}\ \text{m}</math>.	+	<math>\begin{align}
		+	& \mathbf{r}=-24\mathbf{i}+56\mathbf{j}+0\mathbf{k}\ \text{m}
		+	\end{align}</math>
		+
		+	we get the distance from the origin to where it hits the ground is
		+
		+	<math>\sqrt{{{\left( -24 \right)}^{2}}+{{56}^{2}}}=\sqrt{576+3136}=\sqrt{3712}=60\textrm{.}9\ \text{m}</math>

Ian at 14:42, 14 May 2010

2010-05-14T14:42:20Z

←Older revision		Revision as of 14:42, 14 May 2010
Line 9:		Line 9:
	& =-24\mathbf{i}+56\mathbf{j}+0\mathbf{k}\ \text{m} \\		& =-24\mathbf{i}+56\mathbf{j}+0\mathbf{k}\ \text{m} \\
	\end{align}</math>		\end{align}</math>
		+
		+	The point where the frisby was thrown is the origin. This is obtained by putting
		+	<math>t=0</math>
		+	in the expression for
		+	<math>\mathbf{r}</math>
		+	giving <math>\mathbf{r} =0\mathbf{i}+0\mathbf{j}+0\mathbf{k}\ \text{m}</math>.

Ian: New page: We insert $t=0\ \text{s}$ in the expression for the position vector to find the point where the frisby waqs thrown from. $\begin{align} & \mathbf{r}=\left( 5\times 0-{{0...$

2010-04-11T16:24:15Z

New page: We insert <math>t=0\ \text{s}</math> in the expression for the position vector to find the point where the frisby waqs thrown from. <math>\begin{align} & \mathbf{r}=\left( 5\times 0-{{0...

New page

We insert
<math>t=0\ \text{s}</math>
in the expression for the position vector to find the point where the frisby waqs thrown from.

<math>\begin{align}
& \mathbf{r}=\left( 5\times 0-{{0}^{2}} \right)\mathbf{i}+\left( 7\times 0 \right)\mathbf{j}+0\mathbf{k} \\
& \\
& =-24\mathbf{i}+56\mathbf{j}+0\mathbf{k}\ \text{m} \\
\end{align}</math>

Solution 7.4c - Revision history

Ian at 14:56, 14 May 2010

Ian at 14:42, 14 May 2010

Ian: New page: We insert t=0\ \text{s} in the expression for the position vector to find the point where the frisby waqs thrown from. \begin{align} & \mathbf{r}=\left( 5\times 0-{{0...

Ian: New page: We insert $t=0\ \text{s}$ in the expression for the position vector to find the point where the frisby waqs thrown from. $\begin{align} & \mathbf{r}=\left( 5\times 0-{{0...$