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Chapter 3: Problem 3

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of \(340 \mathrm{m} / \mathrm{s}^{2} .\) The ball is launched at an angle of \(51^{\circ}\) above the ground. Determine the horizontal and vertical components of the launch velocity.

Short Answer

Expert verified
The horizontal component is approximately 10.7 m/s, and the vertical component is approximately 13.2 m/s.

Step by step solution

01

Calculate Total Velocity

First, we need to determine the total velocity of the ball at the moment it leaves the kicker's foot. We know the acceleration of the ball and the time it is in contact with the foot. Using the equation: \( v = a \times t \), where \( v \) is the velocity, \( a = 340\, \text{m/s}^2 \) is the acceleration, and \( t = 0.050\, \text{s} \), we obtain \( v = 340 \times 0.050 = 17\, \text{m/s} \). This is the total velocity at launch.
02

Break Down the Velocity into Components

Now, we need to find the horizontal and vertical components of the launch velocity. These can be calculated using trigonometry. We will use the angle of launch, \(51^{\circ}\), to do this.
03

Calculate the Horizontal Component

The horizontal component \( v_x \) of the velocity can be found using the formula: \( v_x = v \times \cos(\theta) \), where \( v = 17\, \text{m/s} \) and \( \theta = 51^{\circ} \). Thus, \( v_x = 17 \times \cos(51^{\circ}) \approx 17 \times 0.6293 \approx 10.7\, \text{m/s} \).
04

Calculate the Vertical Component

Similarly, the vertical component \( v_y \) of the velocity is calculated using: \( v_y = v \times \sin(\theta) \), where \( v = 17\, \text{m/s} \) and \( \theta = 51^{\circ} \). Thus, \( v_y = 17 \times \sin(51^{\circ}) \approx 17 \times 0.7771 \approx 13.2\, \text{m/s} \).

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Key Concepts

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

Horizontal Component of Velocity
When a ball is kicked, it moves in two directions - horizontally and vertically. The horizontal component of velocity, often denoted as \( v_x \), is the part of the velocity that moves parallel to the ground. This component doesn't change during the flight if we ignore air resistance.

To find this horizontal part, we use the formula:
  • \( v_x = v \times \cos(\theta) \)
Here, \( v \) is the total velocity and \( \theta \) is the angle of launch. The cosine function helps us "project" the velocity onto the horizontal axis. In our football example, with a launch velocity of 17 m/s and an angle of 51掳, we multiply by \( \cos(51掳) \approx 0.6293 \).

This gives \( v_x \approx 10.7\) m/s, meaning the ball travels 10.7 meters per second horizontally.
Vertical Component of Velocity
The vertical component of velocity, \( v_y \), is what makes the ball rise and eventually come back down. This part is affected by gravity, which continuously pulls the ball downward.

We calculate \( v_y \) using:
  • \( v_y = v \times \sin(\theta) \)
The sine function helps us "project" the velocity vertically. For a launch speed of 17 m/s at 51掳, we use \( \sin(51掳) \approx 0.7771 \).

So, \( v_y \approx 13.2 \) m/s, meaning the ball ascends at 13.2 meters per second. This helps the ball to reach a certain height and then fall back due to gravity's influence.
Launch Angle
The launch angle, \( \theta \), is crucial in determining how far and high the ball goes. It's the angle between the direction of the initial velocity and the horizontal ground.

This angle affects both the horizontal and vertical components of velocity. A higher angle means more vertical velocity, leading to higher arcs, while a lower angle increases the horizontal reach but decreases the height.

In our example, the launch angle is 51掳. This means the ball strikes a balance, allowing it to travel both a significant height and distance.\( \theta \) determines the trajectory. Adjusting it changes how far and high the projectile can travel, depending on whether you want maximum range or height.

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Most popular questions from this chapter

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?

A fire hose ejects a stream of water at an angle of \(35.0^{\circ}\) above the horizontal. The water leaves the nozzle with a speed of \(25.0 \mathrm{m} / \mathrm{s}\). Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?

A baseball player hits a triple and ends up on third base. A baseball "diamond" is a square, each side of length \(27.4 \mathrm{m},\) with home plate and the three bases on the four corners. What is the magnitude of the player's displacement?

When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of \(10.0 \mathrm{m} / \mathrm{s},\) at an angle of \(31.0^{\circ}\) above the horizontal. (a) What is the range of his leap and (b) for how much time is he in the air?

A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is \(0.870 \mathrm{m} / \mathrm{s}\) at an angle of \(35.0^{\circ}\) above the table, and it lands on the magazine \(0.0770 \mathrm{s}\) after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.
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