Solution 8.1a

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<math>\begin{align}
<math>\begin{align}
& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\
& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\
-
& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\
+
& =40\mathbf{i}+45\mathbf{i}+35\mathbf{j}+400\mathbf{i}+350\mathbf{j} \\ \\
-
& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}
+
& =485\mathbf{i}+385\mathbf{j} \ \text{m}
\end{align}</math>
\end{align}</math>

Revision as of 17:20, 12 April 2010

In this case we have \displaystyle \mathbf{u}=4\mathbf{i} , \displaystyle \mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j} and \displaystyle {{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}.

Substituting these into the equation \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+400\mathbf{i}+350\mathbf{j} gives the position vector of the ball at time \displaystyle 10 as:

\displaystyle \begin{align} & \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\ & =40\mathbf{i}+45\mathbf{i}+35\mathbf{j}+400\mathbf{i}+350\mathbf{j} \\ \\ & =485\mathbf{i}+385\mathbf{j} \ \text{m} \end{align}

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