Question

An arrow is shot from a bow at an angle of 35 degrees above horizontal with an initial speed of 50 m/s what is the arrows horizontal x and y components ?

Answer 1

These kinds of problems can be broken down to a simple right triangle where we want the side lengths.  Knowing the hypotenuse (50) and the angle, we can get the  other sides with trigonometry.  These sides are then the components of the original vector.  I drew it up for you here.

Answer 2

Answer: Horizontal component of arrow :40.955 m/s

Vertical component of arrow :28.675 m/s

Step-by-step explanation:

The initial speed of the arrow = u = 50 m/s

The horizontal component of the arrow =
`u_x=u\cos\theta`

The horizontal component of the arrow =
`u_y=u\sin\theta`

Angle between the velocity vector and x component = 35°

Horizontal component of arrow :

`\cos\theta=(u_x)/(u)`

`\cos35^o=0.8191=(u_x)/(50)`

`u_x=0.8191* 50=40.955 m/s`

Vertical component of arrow :

`\sin\theta=(u_y)/(u)`

`\sin35^o=0.5735=(u_y)/(50)`

`u_y=0.5735* 50=28.675 m/s`

Horizontal component of arrow :40.955 m/s

Vertical component of arrow :28.675 m/s