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Sagot :
To determine the magnitude of the acceleration of the ball, we'll break down the given information and solve it step by step.
### Step 1: Understanding Initial and Final Velocities
- Initial velocity ([tex]\(v_0\)[/tex]) of the ball: [tex]\(2.11 \, \text{m/s}\)[/tex]
- Initial angle ([tex]\(\theta_0\)[/tex]): [tex]\(-37.0^{\circ}\)[/tex]
- Final velocity ([tex]\(v_f\)[/tex]) of the ball: [tex]\(3.80 \, \text{m/s}\)[/tex]
- Final angle ([tex]\(\theta_f\)[/tex]): [tex]\(150.0^{\circ}\)[/tex]
- Contact time ([tex]\(t\)[/tex]): [tex]\(0.19 \, \text{s}\)[/tex]
### Step 2: Convert Angles to Radians
Angles need to be in radians for trigonometric calculations:
- Initial angle in radians: [tex]\(\theta_0 = -37.0^{\circ} = -0.6457718232380625 \, \text{radians}\)[/tex]
- Final angle in radians: [tex]\(\theta_f = 150.0^{\circ} = 2.6179938779914944 \, \text{radians}\)[/tex]
### Step 3: Calculate Initial Velocity Components
Using the initial velocity and angle to find the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] components:
- [tex]\(v_{0_x} = v_0 \cos(\theta_0) = 2.11 \cos(-37.0^{\circ}) = 1.6851209261997877 \, \text{m/s}\)[/tex]
- [tex]\(v_{0_y} = v_0 \sin(\theta_0) = 2.11 \sin(-37.0^{\circ}) = -1.2698296988508218 \, \text{m/s}\)[/tex]
### Step 4: Calculate Final Velocity Components
Using the final velocity and angle to find the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] components:
- [tex]\(v_{f_x} = v_f \cos(\theta_f) = 3.80 \cos(150.0^{\circ}) = -3.290896534380867 \, \text{m/s}\)[/tex]
- [tex]\(v_{f_y} = v_f \sin(\theta_f) = 3.80 \sin(150.0^{\circ}) = 1.8999999999999997 \, \text{m/s}\)[/tex]
### Step 5: Calculate Changes in Velocities
Determine the changes in the velocity components:
- [tex]\(\Delta v_x = v_{f_x} - v_{0_x} = -3.290896534380867 - 1.6851209261997877 = -4.976017460580655 \, \text{m/s}\)[/tex]
- [tex]\(\Delta v_y = v_{f_y} - v_{0_y} = 1.8999999999999997 - (-1.2698296988508218) = 3.1698296988508217 \, \text{m/s}\)[/tex]
### Step 6: Calculate Components of Acceleration
Using the changes in velocity and contact time to determine the acceleration components:
- [tex]\(a_x = \frac{\Delta v_x}{t} = \frac{-4.976017460580655}{0.19} = -26.189565582003446 \, \text{m/s}^2\)[/tex]
- [tex]\(a_y = \frac{\Delta v_y}{t} = \frac{3.1698296988508217}{0.19} = 16.68331420447801 \, \text{m/s}^2\)[/tex]
### Step 7: Calculate the Magnitude of Acceleration
Finally, determine the overall magnitude of the acceleration using the Pythagorean theorem:
- [tex]\(a = \sqrt{a_x^2 + a_y^2}\)[/tex]
- [tex]\(a = \sqrt{(-26.189565582003446)^2 + (16.68331420447801)^2} = 31.051993788151464 \, \text{m/s}^2\)[/tex]
### Answer
The magnitude of the acceleration of the ball is:
[tex]\[ a = 31.051993788151464 \, \text{m/s}^2 \][/tex]
### Step 1: Understanding Initial and Final Velocities
- Initial velocity ([tex]\(v_0\)[/tex]) of the ball: [tex]\(2.11 \, \text{m/s}\)[/tex]
- Initial angle ([tex]\(\theta_0\)[/tex]): [tex]\(-37.0^{\circ}\)[/tex]
- Final velocity ([tex]\(v_f\)[/tex]) of the ball: [tex]\(3.80 \, \text{m/s}\)[/tex]
- Final angle ([tex]\(\theta_f\)[/tex]): [tex]\(150.0^{\circ}\)[/tex]
- Contact time ([tex]\(t\)[/tex]): [tex]\(0.19 \, \text{s}\)[/tex]
### Step 2: Convert Angles to Radians
Angles need to be in radians for trigonometric calculations:
- Initial angle in radians: [tex]\(\theta_0 = -37.0^{\circ} = -0.6457718232380625 \, \text{radians}\)[/tex]
- Final angle in radians: [tex]\(\theta_f = 150.0^{\circ} = 2.6179938779914944 \, \text{radians}\)[/tex]
### Step 3: Calculate Initial Velocity Components
Using the initial velocity and angle to find the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] components:
- [tex]\(v_{0_x} = v_0 \cos(\theta_0) = 2.11 \cos(-37.0^{\circ}) = 1.6851209261997877 \, \text{m/s}\)[/tex]
- [tex]\(v_{0_y} = v_0 \sin(\theta_0) = 2.11 \sin(-37.0^{\circ}) = -1.2698296988508218 \, \text{m/s}\)[/tex]
### Step 4: Calculate Final Velocity Components
Using the final velocity and angle to find the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] components:
- [tex]\(v_{f_x} = v_f \cos(\theta_f) = 3.80 \cos(150.0^{\circ}) = -3.290896534380867 \, \text{m/s}\)[/tex]
- [tex]\(v_{f_y} = v_f \sin(\theta_f) = 3.80 \sin(150.0^{\circ}) = 1.8999999999999997 \, \text{m/s}\)[/tex]
### Step 5: Calculate Changes in Velocities
Determine the changes in the velocity components:
- [tex]\(\Delta v_x = v_{f_x} - v_{0_x} = -3.290896534380867 - 1.6851209261997877 = -4.976017460580655 \, \text{m/s}\)[/tex]
- [tex]\(\Delta v_y = v_{f_y} - v_{0_y} = 1.8999999999999997 - (-1.2698296988508218) = 3.1698296988508217 \, \text{m/s}\)[/tex]
### Step 6: Calculate Components of Acceleration
Using the changes in velocity and contact time to determine the acceleration components:
- [tex]\(a_x = \frac{\Delta v_x}{t} = \frac{-4.976017460580655}{0.19} = -26.189565582003446 \, \text{m/s}^2\)[/tex]
- [tex]\(a_y = \frac{\Delta v_y}{t} = \frac{3.1698296988508217}{0.19} = 16.68331420447801 \, \text{m/s}^2\)[/tex]
### Step 7: Calculate the Magnitude of Acceleration
Finally, determine the overall magnitude of the acceleration using the Pythagorean theorem:
- [tex]\(a = \sqrt{a_x^2 + a_y^2}\)[/tex]
- [tex]\(a = \sqrt{(-26.189565582003446)^2 + (16.68331420447801)^2} = 31.051993788151464 \, \text{m/s}^2\)[/tex]
### Answer
The magnitude of the acceleration of the ball is:
[tex]\[ a = 31.051993788151464 \, \text{m/s}^2 \][/tex]
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