Answer to Question #139301 in Physics for Alfred john Aguilar

Question

A boomerang is thrown with the following displacements: 4.5m east, 7.5m 30° N of W, and 3.0m at 10° S of W. What is the resultant displacement?

Answer 1

Using the following notions,

the vector of the corresponding displacements have the following coordinates:

$$\mathbf{d}_1 = 4.5\times (1,0) = (4.5,0)\\ \mathbf{d}_2 =7.5\times (-\cos30\degree,\sin30\degree) \approx (-6.50,3.75)\\ \mathbf{d}_3 =3\times (-\cos10\degree,-\sin10\degree) \approx (-2.95,-0.52)\\$$

The resultant displacement will be:

$$\mathbf{d}_1 + \mathbf{d}_2+ \mathbf{d}_3 = \\ =(4.5 - 6.5 - 2.95, 0+3.75-0.52) = (-4.95,3.23) \space (m)$$

Answer. $(-4.95,3.23) \space (m)$.