Respuesta :
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: Horizontal component of arrow :40.955 m/s
Vertical component of arrow :28.675 m/s
Explanation:
The initial speed of the arrow = u = 50 m/s
The horizontal component of the arrow =[tex]u_x=u\cos\theta [/tex]
The horizontal component of the arrow =[tex]u_y=u\sin\theta [/tex]
Angle between the velocity vector and x component = 35°
Horizontal component of arrow :
[tex]\cos\theta=\frac{u_x}{u}[/tex]
[tex]\cos35^o=0.8191=\frac{u_x}{50}[/tex]
[tex]u_x=0.8191\times 50=40.955 m/s[/tex]
Vertical component of arrow :
[tex]\sin\theta=\frac{u_y}{u}[/tex]
[tex]\sin35^o=0.5735=\frac{u_y}{50}[/tex]
[tex]u_y=0.5735\times 50=28.675 m/s[/tex]
Horizontal component of arrow :40.955 m/s
Vertical component of arrow :28.675 m/s
