A velocity of 14 m/s is resolved into two components along OA and OB making angles 30^ and…

Question

A velocity of $\dfrac14$ m/s is resolved into two components along $OA$ and $OB$ making angles $30^{\circ}$ and $45^{\circ}$ respectively with the given velocity. Then the component along $OB$ is:

Accepted Answer

Using the resolution formula, component along $OB$ $=\dfrac{v\sin30^{\circ}}{\sin(45^{\circ}+30^{\circ})}$ with $v=\dfrac14$:

$$= \frac{\frac14\cdot\frac12}{\frac{\sqrt3+1}{2\sqrt2}} = \frac{\sqrt6-\sqrt2}8$$

Solutions · 1