Respuesta :
This seems like a calculus problem. I'm assuming you would use cos and sin. so here's the vertical component +10.0m/s multiplied by sin60 = 8.66 rounded to the hundreths place. Now for horizontal, that would be +10.0m/s multiplied by cos60 = 5. hope this helped.
Answer:
Horizontal component: [tex]V_{x}=+5m/s[/tex]
Vertical component: [tex]V_{y}=+8.66m/s[/tex]
Explanation:
We have the launch speed [tex]V =+ 10m/s[/tex] and the angle of 60°, so the components of the speed [tex]V_{x}[/tex] and [tex]V_{y}[/tex] are those shown in the attached image.
[tex]V_{x}[/tex] which is the horizontal component of the velocity can be found by multiplying the cosine of the angle by the initial velocity:
[tex]V_{x}=+10m/s(cos60)\\V_{x}=+10m/s(0.5)\\V_{x}=+5m/s[/tex]
[tex]V_{y}[/tex] which is the verticalcomponent of the velocity is found by multiplying the sine of angle by the initial velocity
[tex]V_{y}=+10m/s(sin60)\\V_{y}=+10m/s(0.866)\\V_{y}=+8.66m/s[/tex]
In summary:
Horizontal component: [tex]V_{x}=+5m/s[/tex]
Vertical component: [tex]V_{y}=+8.66m/s[/tex]
