A ball is thrown with an angle of 25.0 to the horizon with a speed of 18.0 m/s. What are its horizontal and vertical components?

To find the horizontal and vertical components of the initial velocity of the ball, you can use trigonometric functions. The horizontal component (Vx) is given by �⋅cos⁡(�)

V⋅cos(θ), and the vertical component (Vy) is given by �⋅sin⁡(�)

V⋅sin(θ), where:

�

V is the magnitude of the velocity,

�

θ is the angle of the velocity vector with respect to the horizontal.

In this case:

�=18.0 m/s

V=18.0m/s (the magnitude of the velocity),

�=25.0∘

θ=25.0∘

(the angle of the velocity vector).

Horizontal component (��

Vx):

��=�⋅cos⁡(�)

Vx=V⋅cos(θ)

��=18.0 m/s⋅cos⁡(25.0∘)

Vx=18.0m/s⋅cos(25.0∘

)

��≈16.2 m/s

Vx≈16.2m/s

Vertical component (��

Vy):

��=�⋅sin⁡(�)

Vy=V⋅sin(θ)

��=18.0 m/s⋅sin⁡(25.0∘)

Vy=18.0m/s⋅sin(25.0∘

)

��≈7.7 m/s

Vy≈7.7m/s

So, the horizontal component (��

Vx) is approximately 16.2 m/s

16.2m/s, and the vertical component (��

Vy) is approximately 7.7 m/s

7.7m/s.