# An arrow is shot into the air at an elevation angle of 25 degrees. If its projection velocity is 55 m/s. What is the vertical component of its initial velocity? What is its horizontal component?

##### 1 Answer
Jul 4, 2015

Vertical component: 23 m/s
Horizontal component: 50. m/s

#### Explanation:

For a given launch angle $\theta$, the vertical and horizontal components of the initial velocity can be written as

$\left\{\begin{matrix}{v}_{o y} = {v}_{0} \cdot \sin \left(\theta\right) \\ {v}_{0 x} = {v}_{0} \cdot \cos \left(\theta\right)\end{matrix}\right.$

In your case, the two components of the initial speed will be

v_(oy) = "55 m/s" * sin(25^@) = color(green)("23 m/s")

and

v_(0x) = "55 m/s" * cos(25^@) = color(green)("50. m/s")

Both values are rounded to two sig figs, the number of sig figs you gave for the arrow's initial velocity.