## Projectile Motion of a Basketball*A basketball is launched with an initial speed of 8.0 m/s and follows the trajectory shown. The ball enters the basket 0.92s after it is launched. What are the distances x and y?*### Diagram ExplanationThe provided diagram illustrates the trajectory of a basketball launched at an angle of 45 degrees with an initial velocity (v₀) of 8.0 m/s. The diagram shows the following:- The initial launch position of the basketball.- The angle of launch is noted as 45 degrees.- The point where the basketball enters the basket is marked.- The horizontal distance from the launch point to the basket is denoted as 'x.'- The vertical distance from the launch point to the basket is denoted as 'y.'### Understanding the ProblemTo determine the distances 'x' (horizontal distance) and 'y' (vertical distance), we will use the principles of projectile motion. Given the initial speed (v₀), time of flight (0.92 s), and angle of launch (45 degrees), we can apply kinematic equations to find 'x' and 'y.'### Equations and SolutionThe horizontal and vertical components of the initial velocity can be found using trigonometry:- \( v_{0x} = v_0 \cos(45^\circ) \)- \( v_{0y} = v_0 \sin(45^\circ) \)Since \(\cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2}\):- \( v_{0x} = 8.0 \, \text{m/s} \times \frac{\sqrt{2}}{2} = 8.0 \, \text{m/s} \times 0.707 \approx 5.66 \, \text{m/s} \)- \( v_{0y} = 8.0 \, \text{m/s} \times \frac{\sqrt{2}}{2} = 8.0 \, \text{m/s} \times 0.707 \approx 5.66 \, \text{m/s} \)For the horizontal distance (x):- \( x = v_{0x} t \)- \( x = 5.66 \, \text{m/s}

Name: ## Projectile Motion of a Basketball*A basketball is launched with an
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Description: ## Projectile Motion of a Basketball*A basketball is launched with an initial speed of 8.0 m/s and follows the trajectory shown. The ball enters the basket 0.92s after it is launched. What are the distances x and y?*### Diagram ExplanationThe provid

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A basketball is jaunched with an initHal speed of 8.0ms and follow the trajecrony shoun. The ball en ters the bask et 0.9es after it is launched. What are the distances X and Y.

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question

## Projectile Motion of a Basketball

*A basketball is launched with an initial speed of 8.0 m/s and follows the trajectory shown. The ball enters the basket 0.92s after it is launched. What are the distances x and y?*

### Diagram Explanation
The provided diagram illustrates the trajectory of a basketball launched at an angle of 45 degrees with an initial velocity (v₀) of 8.0 m/s. The diagram shows the following:

- The initial launch position of the basketball.
- The angle of launch is noted as 45 degrees.
- The point where the basketball enters the basket is marked.
- The horizontal distance from the launch point to the basket is denoted as 'x.'
- The vertical distance from the launch point to the basket is denoted as 'y.'

### Understanding the Problem

To determine the distances 'x' (horizontal distance) and 'y' (vertical distance), we will use the principles of projectile motion. Given the initial speed (v₀), time of flight (0.92 s), and angle of launch (45 degrees), we can apply kinematic equations to find 'x' and 'y.'

### Equations and Solution

The horizontal and vertical components of the initial velocity can be found using trigonometry:

- \( v_{0x} = v_0 \cos(45^\circ) \)
- \( v_{0y} = v_0 \sin(45^\circ) \)

Since \(\cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2}\):

- \( v_{0x} = 8.0 \, \text{m/s} \times \frac{\sqrt{2}}{2} = 8.0 \, \text{m/s} \times 0.707 \approx 5.66 \, \text{m/s} \)
- \( v_{0y} = 8.0 \, \text{m/s} \times \frac{\sqrt{2}}{2} = 8.0 \, \text{m/s} \times 0.707 \approx 5.66 \, \text{m/s} \)

For the horizontal distance (x):

- \( x = v_{0x} t \)
- \( x = 5.66 \, \text{m/s}

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