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Solution 8.1a

From Mechanics

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Line 12: Line 12:
<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 u=4i , a=09i+07j and r0=400i+350j.

Substituting these into the equation r=ut+21at 2+400i+350j gives the position vector of the ball at time 10 as:

r=(4i)10+21(09i+07j)10 2+400i+350j=40i+45i+35j+400i+350j=485i+385j m

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