Ian at 10:29, 11 April 2010

Ian — Sun, 11 Apr 2010 10:29:23 GMT

←Older revision		Revision as of 10:29, 11 April 2010
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	<math>t=0\ \</math> gives		<math>t=0\ \</math> gives

-	<math>\mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4.9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m}</math>	+	<math>\mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4\textrm{.}9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m}</math>


	<math>t=1\ \</math> gives		<math>t=1\ \</math> gives

-	<math>\mathbf{r}=6\times 1\mathbf{i}+\left( 15\times 1-4.9\times {{1}^{2}} \right)\mathbf{j}=6\mathbf{i}+10.1\mathbf{j}\ \text{m}</math>	+	<math>\mathbf{r}=6\times 1\mathbf{i}+\left( 15\times 1-4\textrm{.}9\times {{1}^{2}} \right)\mathbf{j}=6\mathbf{i}+10\textrm{.}1\mathbf{j}\ \text{m}</math>

	<math>t=2\ \</math> gives		<math>t=2\ \</math> gives

-	<math>\mathbf{r}=6\times 2\mathbf{i}+\left( 15\times 2-4.9\times {{2}^{2}} \right)\mathbf{j}=12\mathbf{i}+10.4\mathbf{j}\ \text{m}</math>	+	<math>\mathbf{r}=6\times 2\mathbf{i}+\left( 15\times 2-4\textrm{.}9\times {{2}^{2}} \right)\mathbf{j}=12\mathbf{i}+10\textrm{.}4\mathbf{j}\ \text{m}</math>
		+
		+
		+	<math>t=3\ \</math> gives
		+
		+	<math>\mathbf{r}=6\times 3\mathbf{i}+\left( 15\times 3-4\textrm{.}9\times {{3}^{2}} \right)\mathbf{j}=18\mathbf{i}+0\textrm{.}9\mathbf{j}\ \text{m}</math>
		+
		+
		+	<math>t=4\ \</math> gives
		+
		+	<math>\mathbf{r}=6\times 4\mathbf{i}+\left( 15\times 4-4\textrm{.}9\times {{4}^{2}} \right)\mathbf{j}=24\mathbf{i}-18\textrm{.}4\mathbf{j}\ \text{m}</math>

Ian: New page: $t=0\ \$ gives $\mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4.9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m}$ $t=1\ \$ gives...

Ian — Sun, 11 Apr 2010 10:24:55 GMT

New page: <math>t=0\ \</math> gives <math>\mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4.9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m}</math> <math>t=1\ \</math> gives...

New page

<math>t=0\ \</math> gives

<math>\mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4.9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m}</math>

<math>t=1\ \</math> gives

<math>\mathbf{r}=6\times 1\mathbf{i}+\left( 15\times 1-4.9\times {{1}^{2}} \right)\mathbf{j}=6\mathbf{i}+10.1\mathbf{j}\ \text{m}</math>

<math>t=2\ \</math> gives

<math>\mathbf{r}=6\times 2\mathbf{i}+\left( 15\times 2-4.9\times {{2}^{2}} \right)\mathbf{j}=12\mathbf{i}+10.4\mathbf{j}\ \text{m}</math>

Solution 7.1a - Revision history

Ian at 10:29, 11 April 2010

Ian: New page: t=0\ \ gives \mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4.9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m} t=1\ \ gives...

Ian: New page: $t=0\ \$ gives $\mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4.9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m}$ $t=1\ \$ gives...