Solution 16.7b
From Mechanics
(Difference between revisions)
(New page: From part a), <math>\mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j}</math> which gives, <math>\begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \alpha =58\textrm{.}0{}^\circ \\ \...) |
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Line 4: | Line 4: | ||
<math>\begin{align} | <math>\begin{align} | ||
- | & \tan \alpha =\frac{4}{2\textrm{.}5} \\ | + | & \tan \alpha =\frac{4}{2\textrm{.}5} \\ |
+ | & \\ | ||
& \alpha =58\textrm{.}0{}^\circ \\ | & \alpha =58\textrm{.}0{}^\circ \\ | ||
\end{align}</math> | \end{align}</math> |
Current revision
From part a),
\displaystyle \mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j} which gives,
\displaystyle \begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \\ & \alpha =58\textrm{.}0{}^\circ \\ \end{align}