Answer :
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]
