Question

Consider a particle with initial velocity v that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

Answer 1

Answer:

v_x = -6.00 m/s

v_y = 10.4 m/s

Explanation:

To calculate the velocity of the x-component we must use the cosine function:

$v_{x} =-v_{0}cos(\alpha )$, where ∝ is the angle 60°, and $v_{0}$ is the velocity 12.0 m/s (note: the negative sign due to the question stating it is "above the negative x-axis).

This gives us a value $v_{x} =-6$ m/s

Similarly for the y-component, we must use the sine function:

$v_{y}=v_{0}sin(\alpha)$

which gives us a value $v_{y} = 10.392...$ ≈ 10.4 m/s (since there are 3 significant figures in the information given in the question, this is rounded to 10.4 m/s).