Ian: New page: In this case we have $\mathbf{u}=5\mathbf{i}$ , $\mathbf{a}=-4\mathbf{j}$ and ${{\mathbf{r}}_{0}}=18\mathbf{i}+14\mathbf{j}$ . Substituting these into ...

2010-04-13T15:05:43Z

New page: In this case we have <math>\mathbf{u}=5\mathbf{i}</math> , <math>\mathbf{a}=-4\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=18\mathbf{i}+14\mathbf{j}</math>. Substituting these into ...

New page

In this case we have
<math>\mathbf{u}=5\mathbf{i}</math>
,
<math>\mathbf{a}=-4\mathbf{j}</math>
and
<math>{{\mathbf{r}}_{0}}=18\mathbf{i}+14\mathbf{j}</math>.

Substituting these into the equation
<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>
gives the position vector of the ball at time <math>t=8 \text{ s }</math> as:

<math>\begin{align}
& \mathbf{r}=(5\mathbf{i})\times 8+\frac{1}{2}(-4\mathbf{j})\times {{8}^{\ 2}}+18\mathbf{i}+14\mathbf{j} \\
& =40\mathbf{i}-128\mathbf{j}+18\mathbf{i}+14\mathbf{j} \\ \\
& =58\mathbf{i}-114\mathbf{j} \ \text{m}
\end{align}</math>

Solution 8.2b - Revision history

Ian: New page: In this case we have \mathbf{u}=5\mathbf{i} , \mathbf{a}=-4\mathbf{j} and {{\mathbf{r}}_{0}}=18\mathbf{i}+14\mathbf{j}. Substituting these into ...

Ian: New page: In this case we have $\mathbf{u}=5\mathbf{i}$ , $\mathbf{a}=-4\mathbf{j}$ and ${{\mathbf{r}}_{0}}=18\mathbf{i}+14\mathbf{j}$ . Substituting these into ...