Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
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]
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
