Question

A soccer player kicks a soccer ball at +10.0 m/s at an angle of 60°. What are the horizontal and vertical components of velocity of the kick?

Answer 1

Answer:

Horizontal component: $V_{x}=+5m/s$

Vertical component: $V_{y}=+8.66m/s$

Explanation:

We have the launch speed $V =+ 10m/s$ and the angle of 60°, so the components of the speed  $V_{x}$ and  $V_{y}$ are those shown in the attached image.

$V_{x}$ which is the horizontal component of the velocity can be found by multiplying the cosine of the angle by the initial velocity:

$V_{x}=+10m/s(cos60)\\V_{x}=+10m/s(0.5)\\V_{x}=+5m/s$

$V_{y}$ which is the verticalcomponent of the velocity is found by multiplying the sine of angle by the initial velocity

$V_{y}=+10m/s(sin60)\\V_{y}=+10m/s(0.866)\\V_{y}=+8.66m/s$

In summary:

Horizontal component: $V_{x}=+5m/s$

Vertical component: $V_{y}=+8.66m/s$

Answer 2

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.