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  \\  \...) | |||
| 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}
 
		  