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To solve for the horizontal and vertical components of the initial velocity of the soccer ball, we need to use trigonometric functions. Here's a detailed, step-by-step solution process:
1. Understand the Problem:
We are given the initial velocity ([tex]\(v_i\)[/tex]) of the soccer ball and the angle ([tex]\(\theta\)[/tex]) at which it is kicked. We need to find the horizontal ([tex]\(v_{ix}\)[/tex]) and vertical ([tex]\(v_{iy}\)[/tex]) components of this initial velocity.
2. Given Data:
- Initial velocity, [tex]\(v_i = 18.0\; \text{m/s}\)[/tex]
- Angle, [tex]\(\theta = 45^\circ\)[/tex] (assuming 45 degrees as the angle for this solution)
3. Trigonometric Relations:
The horizontal and vertical components of the initial velocity can be found using the sine and cosine functions.
- The horizontal component [tex]\(v_{ix}\)[/tex] is given by:
[tex]\[ v_{ix} = v_i \cos(\theta) \][/tex]
- The vertical component [tex]\(v_{iy}\)[/tex] is given by:
[tex]\[ v_{iy} = v_i \sin(\theta) \][/tex]
4. Convert the Angle to Radians:
In trigonometric functions, angles should be in radians. For [tex]\(\theta = 45^\circ\)[/tex]:
[tex]\[ \theta = \frac{45 \times \pi}{180} = \frac{\pi}{4} \; \text{radians} \][/tex]
5. Calculate Horizontal Component:
[tex]\[ v_{ix} = 18.0 \; \text{m/s} \times \cos\left(\frac{\pi}{4}\right) \][/tex]
Since [tex]\(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)[/tex],
[tex]\[ v_{ix} = 18.0 \times \frac{\sqrt{2}}{2} \][/tex]
This simplifies to:
[tex]\[ v_{ix} = 18.0 \times 0.7071 \approx 12.7 \; \text{m/s} \][/tex]
6. Calculate Vertical Component:
[tex]\[ v_{iy} = 18.0 \; \text{m/s} \times \sin\left(\frac{\pi}{4}\right) \][/tex]
Since [tex]\(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)[/tex],
[tex]\[ v_{iy} = 18.0 \times \frac{\sqrt{2}}{2} \][/tex]
This simplifies to:
[tex]\[ v_{iy} = 18.0 \times 0.7071 \approx 12.7 \; \text{m/s} \][/tex]
7. Final Answer:
[tex]\[ \begin{array}{l} v_{i x} = 12.7 \; \text{m/s} \\ v_{i y} = 12.7 \; \text{m/s} \\ \end{array} \][/tex]
So, the horizontal component [tex]\(v_{ix}\)[/tex] is [tex]\(12.7 \; \text{m/s}\)[/tex], and the vertical component [tex]\(v_{iy}\)[/tex] is [tex]\(12.7 \; \text{m/s}\)[/tex].
1. Understand the Problem:
We are given the initial velocity ([tex]\(v_i\)[/tex]) of the soccer ball and the angle ([tex]\(\theta\)[/tex]) at which it is kicked. We need to find the horizontal ([tex]\(v_{ix}\)[/tex]) and vertical ([tex]\(v_{iy}\)[/tex]) components of this initial velocity.
2. Given Data:
- Initial velocity, [tex]\(v_i = 18.0\; \text{m/s}\)[/tex]
- Angle, [tex]\(\theta = 45^\circ\)[/tex] (assuming 45 degrees as the angle for this solution)
3. Trigonometric Relations:
The horizontal and vertical components of the initial velocity can be found using the sine and cosine functions.
- The horizontal component [tex]\(v_{ix}\)[/tex] is given by:
[tex]\[ v_{ix} = v_i \cos(\theta) \][/tex]
- The vertical component [tex]\(v_{iy}\)[/tex] is given by:
[tex]\[ v_{iy} = v_i \sin(\theta) \][/tex]
4. Convert the Angle to Radians:
In trigonometric functions, angles should be in radians. For [tex]\(\theta = 45^\circ\)[/tex]:
[tex]\[ \theta = \frac{45 \times \pi}{180} = \frac{\pi}{4} \; \text{radians} \][/tex]
5. Calculate Horizontal Component:
[tex]\[ v_{ix} = 18.0 \; \text{m/s} \times \cos\left(\frac{\pi}{4}\right) \][/tex]
Since [tex]\(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)[/tex],
[tex]\[ v_{ix} = 18.0 \times \frac{\sqrt{2}}{2} \][/tex]
This simplifies to:
[tex]\[ v_{ix} = 18.0 \times 0.7071 \approx 12.7 \; \text{m/s} \][/tex]
6. Calculate Vertical Component:
[tex]\[ v_{iy} = 18.0 \; \text{m/s} \times \sin\left(\frac{\pi}{4}\right) \][/tex]
Since [tex]\(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)[/tex],
[tex]\[ v_{iy} = 18.0 \times \frac{\sqrt{2}}{2} \][/tex]
This simplifies to:
[tex]\[ v_{iy} = 18.0 \times 0.7071 \approx 12.7 \; \text{m/s} \][/tex]
7. Final Answer:
[tex]\[ \begin{array}{l} v_{i x} = 12.7 \; \text{m/s} \\ v_{i y} = 12.7 \; \text{m/s} \\ \end{array} \][/tex]
So, the horizontal component [tex]\(v_{ix}\)[/tex] is [tex]\(12.7 \; \text{m/s}\)[/tex], and the vertical component [tex]\(v_{iy}\)[/tex] is [tex]\(12.7 \; \text{m/s}\)[/tex].
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