Answer :
Sure! Let's find the horizontal and vertical components of the soccer ball's velocity step-by-step.
1. Understand the problem: We have a soccer ball kicked at a speed of 10 m/s at an angle of 60 degrees above the horizontal. We need to find the horizontal (V_x) and vertical (V_y) components of this velocity.
2. Use trigonometric functions:
- The horizontal component of the velocity (V_x) can be found using the cosine function.
- The vertical component of the velocity (V_y) can be found using the sine function.
3. Calculate the horizontal component (V_x):
- Use the cosine of the angle to calculate V_x.
- Formula: [tex]\( V_x = V \cdot \cos(\theta) \)[/tex]
- Here, [tex]\( V = 10 \, \text{m/s} \)[/tex] and [tex]\( \theta = 60^\circ \)[/tex].
- So, [tex]\( V_x = 10 \cdot \cos(60^\circ) \)[/tex].
4. Calculate the vertical component (V_y):
- Use the sine of the angle to calculate V_y.
- Formula: [tex]\( V_y = V \cdot \sin(\theta) \)[/tex]
- So, [tex]\( V_y = 10 \cdot \sin(60^\circ) \)[/tex].
5. Get the values from trigonometric tables:
- [tex]\( \cos(60^\circ) = 0.5 \)[/tex]
- [tex]\( \sin(60^\circ) \approx 0.866 \)[/tex]
6. Substitute these values into the formulas:
- [tex]\( V_x = 10 \cdot 0.5 = 5.00 \, \text{m/s} \)[/tex]
- [tex]\( V_y = 10 \cdot 0.866 = 8.66 \, \text{m/s} \)[/tex]
Therefore, the correct horizontal and vertical components of the velocity are [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex] and [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
This matches the choice: [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex] and [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
1. Understand the problem: We have a soccer ball kicked at a speed of 10 m/s at an angle of 60 degrees above the horizontal. We need to find the horizontal (V_x) and vertical (V_y) components of this velocity.
2. Use trigonometric functions:
- The horizontal component of the velocity (V_x) can be found using the cosine function.
- The vertical component of the velocity (V_y) can be found using the sine function.
3. Calculate the horizontal component (V_x):
- Use the cosine of the angle to calculate V_x.
- Formula: [tex]\( V_x = V \cdot \cos(\theta) \)[/tex]
- Here, [tex]\( V = 10 \, \text{m/s} \)[/tex] and [tex]\( \theta = 60^\circ \)[/tex].
- So, [tex]\( V_x = 10 \cdot \cos(60^\circ) \)[/tex].
4. Calculate the vertical component (V_y):
- Use the sine of the angle to calculate V_y.
- Formula: [tex]\( V_y = V \cdot \sin(\theta) \)[/tex]
- So, [tex]\( V_y = 10 \cdot \sin(60^\circ) \)[/tex].
5. Get the values from trigonometric tables:
- [tex]\( \cos(60^\circ) = 0.5 \)[/tex]
- [tex]\( \sin(60^\circ) \approx 0.866 \)[/tex]
6. Substitute these values into the formulas:
- [tex]\( V_x = 10 \cdot 0.5 = 5.00 \, \text{m/s} \)[/tex]
- [tex]\( V_y = 10 \cdot 0.866 = 8.66 \, \text{m/s} \)[/tex]
Therefore, the correct horizontal and vertical components of the velocity are [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex] and [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
This matches the choice: [tex]\( V_x = +5.00 \, \text{m/s} \)[/tex] and [tex]\( V_y = +8.66 \, \text{m/s} \)[/tex].
