To find the direction of the acceleration of the ball after being hit by the bat, we need to follow these steps:
### 1. Determine the initial and final velocity components
We start by finding the x and y components of both the initial and final velocities.
Initial Velocity (before the hit):
- Magnitude: [tex]\(v_1 = 2.15 \, \text{m/s}\)[/tex]
- Angle: [tex]\(\theta_1 = -72.0^\circ\)[/tex]
To find the components:
[tex]\[
v_{1x} = v_1 \cos(\theta_1) = 2.15 \cos(-72.0^\circ) \approx 0.664 \, \text{m/s}
\][/tex]
[tex]\[
v_{1y} = v_1 \sin(\theta_1) = 2.15 \sin(-72.0^\circ) \approx -2.045 \, \text{m/s}
\][/tex]
Final Velocity (after the hit):
- Magnitude: [tex]\(v_2 = 3.12 \, \text{m/s}\)[/tex]
- Angle: [tex]\(\theta_2 = 135.0^\circ\)[/tex]
To find the components:
[tex]\[
v_{2x} = v_2 \cos(\theta_2) = 3.12 \cos(135.0^\circ) \approx -2.206 \, \text{m/s}
\][/tex]
[tex]\[
v_{2y} = v_2 \sin(\theta_2) = 3.12 \sin(135.0^\circ) \approx 2.206 \, \text{m/s}
\][/tex]
### 2. Calculate the change in velocity components
Next, we find the change in the velocity components (i.e., the differences between the final and initial components):
[tex]\[
\Delta v_x = v_{2x} - v_{1x} = -2.206 - 0.664 \approx -2.871 \, \text{m/s}
\][/tex]
[tex]\[
\Delta v_y = v_{2y} - v_{1y} = 2.206 - (-2.045) \approx 4.251 \, \text{m/s}
\][/tex]
### 3. Determine the acceleration components
Acceleration [tex]\(a_x\)[/tex] and [tex]\(a_y\)[/tex] can be obtained by dividing these changes in velocity by the time of contact [tex]\( t = 0.28 \, \text{s} \)[/tex]:
[tex]\[
a_x = \frac{\Delta v_x}{t} = \frac{-2.871}{0.28} \approx -10.252 \, \text{m/s}^2
\][/tex]
[tex]\[
a_y = \frac{\Delta v_y}{t} = \frac{4.251}{0.28} \approx 15.182 \, \text{m/s}^2
\][/tex]
### 4. Calculate the direction of the acceleration
Finally, we find the direction (angle) [tex]\(\theta_{\text{acc}}\)[/tex] of the acceleration using the tangent function:
[tex]\[
\theta_{\text{acc}} = \tan^{-2} \left(\frac{a_y}{a_x}\right) = \tan^{-1}\left(\frac{15.182}{-10.252}\right)
\][/tex]
This gives us:
[tex]\[
\theta_{\text{acc}} \approx 124.03^\circ
\][/tex]
So, the direction of the acceleration of the ball is approximately [tex]\( 124.03^\circ \)[/tex].
