Solution 8.2b
From Mechanics
(Difference between revisions)
(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 ...) |
Current revision
In this case we have \displaystyle \mathbf{u}=5\mathbf{i} , \displaystyle \mathbf{a}=-4\mathbf{j} and \displaystyle {{\mathbf{r}}_{0}}=18\mathbf{i}+14\mathbf{j}.
Substituting these into the equation \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} gives the position vector of the ball at time \displaystyle t=8 \text{ s } as:
\displaystyle \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}
