Question

a vector has a magnitude of 50 in the 35º direction. what are the horizontal and vertical components of this vector?

Answer 1

Answer:

Horizontal components = 40.95

Vertical components = 28.68

Step-by-step explanation:

We have to find the horizontal and vertical components of the vector given magnitude and direction of the vector.

To find the horizontal and vertical components the formula used is:

Vector u =( || u || cos θ , || u || sin θ )

where u = 50

and θ = 35

putting the values

Vector u =( 50 * cos (35) , 50 * sin (35))

              = (50 * 0.8191 , 50 * 0.5735)

             =(40.95,28.68)

so horizontal components = 40.95

vertical components = 28.68

Answer 2

Answer:

Horizontal component: 40.95

Vertical component: 28.67

Step-by-step explanation:

To find the horizontal and the vertical component of this vector, you need:

The formula for calculate the horizontal component:

$a_x=|a|cos\alpha$

Where $|a|$ is the magnitude

The formula for calculate the vertical component:

$a_y=|a|sin\alpha$

Where $|a|$ is the magnitude

Given the angle 35 degrees and the magnitude,you can substitute values into the formulas.

Therefore, you get  that the horizontal component and the vertical component of this vector are:

$a_x=50cos(35\°)=40.95$

$a_y=50sin(35\°)=28.67$