Solution 7.1a

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(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...)
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Revision as of 10:24, 11 April 2010

\displaystyle t=0\ \ gives

\displaystyle \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}


\displaystyle t=1\ \ gives

\displaystyle \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}

\displaystyle t=2\ \ gives

\displaystyle \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}

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