Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To find the magnitude of the acceleration of the ball, let's break down the problem into step-by-step calculations.
### Step 1: Determine the Initial and Final Velocity Components
First, we separate the initial and final velocities into their x and y components. Let's start by converting the angles from degrees to radians:
1. Initial angle: [tex]\(-45.0^\circ\)[/tex]
2. Final angle: [tex]\(125.0^\circ\)[/tex]
### Conversion to Radians
We'll need to convert these angles to radians:
[tex]\[ \text{angle\_initial\_rad} = -45.0^\circ \times \frac{\pi}{180} \approx -0.7854 \text{ radians} \][/tex]
[tex]\[ \text{angle\_final\_rad} = 125.0^\circ \times \frac{\pi}{180} \approx 2.1817 \text{ radians} \][/tex]
### Step 2: Compute Initial and Final Velocity Components
Using trigonometric functions:
- Initial velocity components:
[tex]\[ v_{\text{initial\_x}} = 1.75 \, \text{m/s} \times \cos(-0.7854) \approx 1.2374 \, \text{m/s} \][/tex]
[tex]\[ v_{\text{initial\_y}} = 1.75 \, \text{m/s} \times \sin(-0.7854) \approx -1.2374 \, \text{m/s} \][/tex]
- Final velocity components:
[tex]\[ v_{\text{final\_x}} = 2.95 \, \text{m/s} \times \cos(2.1817) \approx -1.6921 \, \text{m/s} \][/tex]
[tex]\[ v_{\text{final\_y}} = 2.95 \, \text{m/s} \times \sin(2.1817) \approx 2.4165 \, \text{m/s} \][/tex]
### Step 3: Calculate the Change in Velocity Components
- Change in the [tex]\(x\)[/tex]-component of velocity:
[tex]\[ \Delta v_x = v_{\text{final\_x}} - v_{\text{initial\_x}} \approx -1.6921 - 1.2374 \approx -2.9295 \, \text{m/s} \][/tex]
- Change in the [tex]\(y\)[/tex]-component of velocity:
[tex]\[ \Delta v_y = v_{\text{final\_y}} - v_{\text{initial\_y}} \approx 2.4165 - (-1.2374) \approx 3.6539 \, \text{m/s} \][/tex]
### Step 4: Compute the Magnitude of the Change in Velocity
The magnitude of the change in velocity ([tex]\(\Delta v\)[/tex]) can be calculated using Pythagoras' theorem:
[tex]\[ \Delta v = \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2} \][/tex]
[tex]\[ \Delta v \approx \sqrt{(-2.9295)^2 + (3.6539)^2} \approx \sqrt{8.5823 + 13.3539} \approx \sqrt{21.9362} \approx 4.6833 \, \text{m/s} \][/tex]
### Step 5: Calculate the Magnitude of the Acceleration
Finally, to find the magnitude of the acceleration ([tex]\(a\)[/tex]), we use the formula:
[tex]\[ a = \frac{\Delta v}{\text{contact time}} \][/tex]
[tex]\[ a \approx \frac{4.6833 \, \text{m/s}}{0.30 \, \text{s}} \approx 15.6109 \, \text{m/s}^2 \][/tex]
### Conclusion
The magnitude of the acceleration of the ball is:
[tex]\[ a \approx 15.6109 \, \text{m/s}^2 \][/tex]
### Step 1: Determine the Initial and Final Velocity Components
First, we separate the initial and final velocities into their x and y components. Let's start by converting the angles from degrees to radians:
1. Initial angle: [tex]\(-45.0^\circ\)[/tex]
2. Final angle: [tex]\(125.0^\circ\)[/tex]
### Conversion to Radians
We'll need to convert these angles to radians:
[tex]\[ \text{angle\_initial\_rad} = -45.0^\circ \times \frac{\pi}{180} \approx -0.7854 \text{ radians} \][/tex]
[tex]\[ \text{angle\_final\_rad} = 125.0^\circ \times \frac{\pi}{180} \approx 2.1817 \text{ radians} \][/tex]
### Step 2: Compute Initial and Final Velocity Components
Using trigonometric functions:
- Initial velocity components:
[tex]\[ v_{\text{initial\_x}} = 1.75 \, \text{m/s} \times \cos(-0.7854) \approx 1.2374 \, \text{m/s} \][/tex]
[tex]\[ v_{\text{initial\_y}} = 1.75 \, \text{m/s} \times \sin(-0.7854) \approx -1.2374 \, \text{m/s} \][/tex]
- Final velocity components:
[tex]\[ v_{\text{final\_x}} = 2.95 \, \text{m/s} \times \cos(2.1817) \approx -1.6921 \, \text{m/s} \][/tex]
[tex]\[ v_{\text{final\_y}} = 2.95 \, \text{m/s} \times \sin(2.1817) \approx 2.4165 \, \text{m/s} \][/tex]
### Step 3: Calculate the Change in Velocity Components
- Change in the [tex]\(x\)[/tex]-component of velocity:
[tex]\[ \Delta v_x = v_{\text{final\_x}} - v_{\text{initial\_x}} \approx -1.6921 - 1.2374 \approx -2.9295 \, \text{m/s} \][/tex]
- Change in the [tex]\(y\)[/tex]-component of velocity:
[tex]\[ \Delta v_y = v_{\text{final\_y}} - v_{\text{initial\_y}} \approx 2.4165 - (-1.2374) \approx 3.6539 \, \text{m/s} \][/tex]
### Step 4: Compute the Magnitude of the Change in Velocity
The magnitude of the change in velocity ([tex]\(\Delta v\)[/tex]) can be calculated using Pythagoras' theorem:
[tex]\[ \Delta v = \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2} \][/tex]
[tex]\[ \Delta v \approx \sqrt{(-2.9295)^2 + (3.6539)^2} \approx \sqrt{8.5823 + 13.3539} \approx \sqrt{21.9362} \approx 4.6833 \, \text{m/s} \][/tex]
### Step 5: Calculate the Magnitude of the Acceleration
Finally, to find the magnitude of the acceleration ([tex]\(a\)[/tex]), we use the formula:
[tex]\[ a = \frac{\Delta v}{\text{contact time}} \][/tex]
[tex]\[ a \approx \frac{4.6833 \, \text{m/s}}{0.30 \, \text{s}} \approx 15.6109 \, \text{m/s}^2 \][/tex]
### Conclusion
The magnitude of the acceleration of the ball is:
[tex]\[ a \approx 15.6109 \, \text{m/s}^2 \][/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
