To determine the direction of the acceleration of the ball, we need to follow a series of steps that involve resolving the velocities into their components, calculating the accelerations in the x and y directions, and then determining the direction of the acceleration vector.
### Step 1: Calculate initial and final velocity components
Initial Velocity in Components:
Given:
- Initial velocity ([tex]\(v_i\)[/tex]) = 1.13 m/s
- Initial angle ([tex]\(\theta_i\)[/tex]) = -20.5°
To find the x and y components of the initial velocity ([tex]\(v_{ix}\)[/tex], [tex]\(v_{iy}\)[/tex]):
[tex]\[
v_{ix} = v_i \cos(\theta_i)
\][/tex]
[tex]\[
v_{iy} = v_i \sin(\theta_i)
\][/tex]
Using the values:
[tex]\[
v_{ix} = 1.13 \cos(-20.5°) = 1.058439573850689 \, \text{m/s}
\][/tex]
[tex]\[
v_{iy} = 1.13 \sin(-20.5°) = -0.39573434082319814 \, \text{m/s}
\][/tex]
Final Velocity in Components:
Given:
- Final velocity ([tex]\(v_f\)[/tex]) = 2.10 m/s
- Final angle ([tex]\(\theta_f\)[/tex]) = 142°
To find the x and y components of the final velocity ([tex]\(v_{fx}\)[/tex], [tex]\(v_{fy}\)[/tex]):
[tex]\[
v_{fx} = v_f \cos(\theta_f)
\][/tex]
[tex]\[
v_{fy} = v_f \sin(\theta_f)
\][/tex]
Using the values:
[tex]\[
v_{fx} = 2.10 \cos(142°) = -1.654822582574116 \, \text{m/s}
\][/tex]
[tex]\[
v_{fy} = 2.10 \sin(142°) = 1.2928890981838828 \, \text{m/s}
\][/tex]
### Step 2: Calculate the acceleration components
Given:
- Contact time ([tex]\(t\)[/tex]) = 0.25 s
The acceleration components ([tex]\(a_x\)[/tex], [tex]\(a_y\)[/tex]) can be calculated using the change in velocity over the contact time:
[tex]\[
a_x = \frac{v_{fx} - v_{ix}}{t}
\][/tex]
[tex]\[
a_y = \frac{v_{fy} - v_{iy}}{t}
\][/tex]
Using the values:
[tex]\[
a_x = \frac{-1.654822582574116 - 1.058439573850689}{0.25} = -10.85304862569922 \, \text{m/s}^2
\][/tex]
[tex]\[
a_y = \frac{1.2928890981838828 + 0.39573434082319814}{0.25} = 6.754493756028324 \, \text{m/s}^2
\][/tex]
### Step 3: Calculate the direction of the acceleration
The direction of the acceleration ([tex]\(\theta_a\)[/tex]) can be found using the arctangent function:
[tex]\[
\theta_a = \arctan \left(\frac{a_y}{a_x}\right)
\][/tex]
Substitute the values:
[tex]\[
\theta_a = \arctan \left(\frac{6.754493756028324}{-10.85304862569922}\right) = \arctan(-0.622264331) = -32.311^\circ
\][/tex]
Since the angle is negative, we add [tex]\(180^\circ\)[/tex] to ensure it falls within the conventional range for direction of acceleration:
[tex]\[
\theta_a + 180^\circ = -32.311^\circ + 180^\circ = 148.103^\circ
\][/tex]
Therefore, the direction of the acceleration of the ball is:
[tex]\[
\boxed{148.103^\circ}
\][/tex]
