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A soccer player kicks the ball toward a goal that is \(16.8 \mathrm{m}\) in front of him. The ball leaves his foot at a speed of \(16.0 \mathrm{m} / \mathrm{s}\) and an angle of \(28.0^{\circ}\) above the ground. Find the speed of the ball when the goalie catches it in front of the net.

Short Answer

Expert verified
The ball's speed when caught by the goalie is approximately 16 m/s.

Step by step solution

01

Determine Vertical and Horizontal Components

First, we need to find the vertical and horizontal components of the initial velocity. The horizontal component \(v_{0x}\) is calculated using \(v_{0x} = v_0 \cos(\theta)\), and the vertical component \(v_{0y}\) using \(v_{0y} = v_0 \sin(\theta)\).Given, \(v_0 = 16.0 \, \text{m/s}\) and \(\theta = 28.0^{\circ}\), calculate:- \(v_{0x} = 16.0 \, \cos(28.0^{\circ})\)- \(v_{0y} = 16.0 \, \sin(28.0^{\circ})\)
02

Calculate Horizontal Velocity

The horizontal velocity \(v_x\) remains constant throughout the ball's flight because there is no horizontal acceleration. Thus, \(v_x = v_{0x}\). Identify this value from Step 1.
03

Calculate Final Vertical Velocity

The final vertical velocity \(v_{y}\) can be calculated using the kinematic equation considering the initial vertical velocity, gravitational acceleration, and time.\[ v_{y} = v_{0y} - g \, t \]Where \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity). We will find the time \(t\) in the next step.
04

Calculate Time of Flight

To find time \(t\), use the horizontal motion equation:\[ x = v_{0x} \, t \]Where \(x = 16.8 \, \text{m}\) (the distance to the goal). Solving for \(t\):\[ t = \frac{x}{v_{0x}} \]Substitute \(x\) and \(v_{0x}\) from earlier steps to find \(t\).
05

Calculate Speed of Ball When Caught

The ball's speed when caught, \(v\), is determined by the Pythagorean theorem, combining horizontal and vertical components:\[ v = \sqrt{v_x^2 + v_y^2} \]Using the known \(v_x\) and the calculated \(v_y\) from previous steps, substitute and solve to find the ball's speed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal and Vertical Components
To understand projectile motion, we first need to break it down into horizontal and vertical components. When an object, like a soccer ball, is kicked at an angle, its movement can be understood by separating it into these two directions. This helps in analyzing the path and velocity of the ball more easily.

The horizontal component of velocity, denoted as \( v_{0x} \), can be found using the equation \( v_{0x} = v_0 \cos(\theta) \). Here, \( v_0 \) represents the initial speed of the ball, and \( \theta \) is the angle at which the ball is kicked. In the exercise, for instance, \( v_0 \) is given as \( 16.0 \, m/s \) and \( \theta \) is \( 28.0^\circ \).

The vertical component \( v_{0y} \) is found similarly: \( v_{0y} = v_0 \sin(\theta) \). This equation allows us to compute the upward velocity of the ball initially. By using trigonometric functions, we break the velocity into understandable parts, making it easier to apply further physics calculations.
Kinematic Equations
Once we have our components, we look at kinematic equations to track the motion over time. Kinematic equations are powerful tools in physics that predict future movement of an object under constant acceleration, like gravity for our soccer ball.

The vertical motion changes due to gravity, which acts downwards, typically at \( 9.81 \, \text{m/s}^2 \). We can calculate the final vertical velocity \( v_{y} \) as the ball falls using the formula:
  • \( v_{y} = v_{0y} - g \, t \)
where \( g \) is the acceleration due to gravity and \( t \) is the time.

For horizontal motion, since no forces like friction or air resistance are considered here, the horizontal velocity \( v_x \) remains constant. The kinematic equation for horizontal motion becomes:
  • \( x = v_{0x} \, t \)
helping us find time of travel if \( x \), the horizontal distance, is known.
Time of Flight
"Time of flight" refers to the total time an object stays in the air. In projectile motion, calculating this time can help us understand how long it takes for the soccer ball to travel from the player's foot to the goal.

For horizontal distance, the time of flight can be calculated from the equation:
  • \( t = \frac{x}{v_{0x}} \)
where \( x \) is the distance to the goal, and \( v_{0x} \) is the horizontal component of the initial velocity. In our soccer example, the goal is \( 16.8 \, \text{m} \) away.

This calculation directly influences other aspects of the ball's motion, like the change in vertical velocity over time. Hence, understanding and calculating the time of flight is crucial to piecing together the full motion picture of any projectile, such as the soccer ball in the exercise.

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Most popular questions from this chapter

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be \(162 \mathrm{km}\) away. The radar antenna is pointing upward at an angle of \(62.3^{\circ}\) from the ground. Find the \(x\) and \(y\) components (in \(\mathrm{km}\) ) of the position vector of the satellite, relative to the antenna.

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced \(51.0 \mathrm{m}\) horizontally from the end of the ramp. His velocity, just before landing, is \(23.0 \mathrm{m} / \mathrm{s}\) and points in a direction \(43.0^{\circ}\) below the horizontal. Neglecting air resistance and any lift he experiences while airborne, find his initial velocity (magnitude and direction) when he left the end of the ramp. Express the direction as an angle relative to the horizontal.

The captain of a plane wishes to proceed due west. The cruising speed of the plane is \(245 \mathrm{m} / \mathrm{s}\) relative to the air. A weather report indicates that a \(38.0-\mathrm{m} / \mathrm{s}\) wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane?

In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is \(35.0 \mathrm{m},\) and the heights of the Jackson and Walton buildings are, respectively, \(15.0 \mathrm{m}\) and \(22.0 \mathrm{m} .\) Ignore air resistance. (a) The first balloon launched by the Jackson team hits Walton dorm \(2.0 \mathrm{s}\) after launch, striking it halfway between the ground and the roof. Find the direction of the balloon's initial velocity. Give your answer as an angle measured above the horizontal. (b) A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed of this second balloon.

A skateboarder shoots off a ramp with a velocity of \(6.6 \mathrm{m} / \mathrm{s}\), directed at an angle of \(58^{\circ}\) above the horizontal. The end of the ramp is \(1.2 \mathrm{m}\) above the ground. Let the \(x\) axis be parallel to the ground, the \(+y\) direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

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