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Answer :
To find the horizontal and vertical components of the velocity when a soccer ball is kicked at an angle, we need to break down the original velocity into its components using trigonometry.
Here's a step-by-step solution:
1. Identify the Given Values:
- The initial velocity (speed of the kick) is [tex]\( 10 \, \text{m/s} \)[/tex].
- The angle of the kick is [tex]\( 60^\circ \)[/tex].
2. Understand the Components:
- The horizontal component ([tex]\( V_x \)[/tex]) is found using the cosine of the angle.
- The vertical component ([tex]\( V_y \)[/tex]) is found using the sine of the angle.
3. Calculate the Horizontal Component ([tex]\( V_x \)[/tex]):
- Use the formula:
[tex]\[
V_x = \text{velocity} \times \cos(\text{angle})
\][/tex]
- Plug in the values:
[tex]\[
V_x = 10 \times \cos(60^\circ)
\][/tex]
- Since [tex]\( \cos(60^\circ) = 0.5 \)[/tex],
[tex]\[
V_x = 10 \times 0.5 = 5.0 \, \text{m/s}
\][/tex]
4. Calculate the Vertical Component ([tex]\( V_y \)[/tex]):
- Use the formula:
[tex]\[
V_y = \text{velocity} \times \sin(\text{angle})
\][/tex]
- Plug in the values:
[tex]\[
V_y = 10 \times \sin(60^\circ)
\][/tex]
- Since [tex]\( \sin(60^\circ) = 0.866 \)[/tex],
[tex]\[
V_y = 10 \times 0.866 = 8.66 \, \text{m/s}
\][/tex]
5. Conclusion:
- The horizontal component of the velocity is [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex].
- The vertical component of the velocity is [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
Therefore, the correct option is:
- [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex] and [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
Here's a step-by-step solution:
1. Identify the Given Values:
- The initial velocity (speed of the kick) is [tex]\( 10 \, \text{m/s} \)[/tex].
- The angle of the kick is [tex]\( 60^\circ \)[/tex].
2. Understand the Components:
- The horizontal component ([tex]\( V_x \)[/tex]) is found using the cosine of the angle.
- The vertical component ([tex]\( V_y \)[/tex]) is found using the sine of the angle.
3. Calculate the Horizontal Component ([tex]\( V_x \)[/tex]):
- Use the formula:
[tex]\[
V_x = \text{velocity} \times \cos(\text{angle})
\][/tex]
- Plug in the values:
[tex]\[
V_x = 10 \times \cos(60^\circ)
\][/tex]
- Since [tex]\( \cos(60^\circ) = 0.5 \)[/tex],
[tex]\[
V_x = 10 \times 0.5 = 5.0 \, \text{m/s}
\][/tex]
4. Calculate the Vertical Component ([tex]\( V_y \)[/tex]):
- Use the formula:
[tex]\[
V_y = \text{velocity} \times \sin(\text{angle})
\][/tex]
- Plug in the values:
[tex]\[
V_y = 10 \times \sin(60^\circ)
\][/tex]
- Since [tex]\( \sin(60^\circ) = 0.866 \)[/tex],
[tex]\[
V_y = 10 \times 0.866 = 8.66 \, \text{m/s}
\][/tex]
5. Conclusion:
- The horizontal component of the velocity is [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex].
- The vertical component of the velocity is [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
Therefore, the correct option is:
- [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex] and [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
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