To determine the magnitude of the acceleration of the ball, we need to follow a series of steps involving vector components, changes in velocity, and ultimately calculating the acceleration.
Here’s a step-by-step breakdown:
### 1. Initial Velocity Components
First, we need to resolve the initial velocity of the ball ([tex]\(1.13 \, \text{m/s}\)[/tex]) at an angle of [tex]\(-20.5^{\circ}\)[/tex] into its horizontal (x) and vertical (y) components.
[tex]\[ v_{ix} = 1.13 \cos(-20.5^\circ) \][/tex]
[tex]\[ v_{iy} = 1.13 \sin(-20.5^\circ) \][/tex]
The components are:
[tex]\[ v_{ix} \approx 1.0584 \, \text{m/s} \][/tex]
[tex]\[ v_{iy} \approx -0.3957 \, \text{m/s} \][/tex]
### 2. Final Velocity Components
Next, we resolve the final velocity of the ball ([tex]\(2.10 \, \text{m/s}\)[/tex]) at an angle of [tex]\(142^{\circ}\)[/tex] into its horizontal (x) and vertical (y) components.
[tex]\[ v_{fx} = 2.10 \cos(142^\circ) \][/tex]
[tex]\[ v_{fy} = 2.10 \sin(142^\circ) \][/tex]
The components are:
[tex]\[ v_{fx} \approx -1.6548 \, \text{m/s} \][/tex]
[tex]\[ v_{fy} \approx 1.2929 \, \text{m/s} \][/tex]
### 3. Change in Velocity Components
Now, we calculate the change in the horizontal and vertical components of the velocity ([tex]\(\Delta v_x\)[/tex] and [tex]\(\Delta v_y\)[/tex]).
[tex]\[ \Delta v_x = v_{fx} - v_{ix} \][/tex]
[tex]\[ \Delta v_y = v_{fy} - v_{iy} \][/tex]
The changes in the components are:
[tex]\[ \Delta v_x \approx -1.6548 \, \text{m/s} - 1.0584 \, \text{m/s} \approx -2.7133 \, \text{m/s} \][/tex]
[tex]\[ \Delta v_y \approx 1.2929 \, \text{m/s} - (-0.3957 \, \text{m/s}) \approx 1.6886 \, \text{m/s} \][/tex]
### 4. Magnitude of the Change in Velocity
To find the magnitude of the change in velocity, [tex]\(\Delta v\)[/tex], we use the Pythagorean theorem:
[tex]\[ \Delta v = \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2} \][/tex]
Substitute in the changes in velocity:
[tex]\[ \Delta v \approx \sqrt{(-2.7133 \, \text{m/s})^2 + (1.6886 \, \text{m/s})^2} \approx 3.1958 \, \text{m/s} \][/tex]
### 5. Magnitude of the Acceleration
Finally, to find the acceleration, we divide the magnitude of the change in velocity by the contact time ([tex]\( \text{contact time} = 0.25 \, \text{s} \)[/tex]):
[tex]\[ a = \frac{\Delta v}{\text{contact time}} \][/tex]
Substitute the values:
[tex]\[ a \approx \frac{3.1958 \, \text{m/s}}{0.25 \, \text{s}} \approx 12.7833 \, \text{m/s}^2 \][/tex]
Hence, the magnitude of the acceleration of the ball is approximately:
[tex]\[ a \approx 12.78 \, \text{m/s}^2 \][/tex]
