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Chapter 3: Problem 13

A volleyball is spiked so that it has an initial velocity of \(15 \mathrm{m} / \mathrm{s}\) directed downward at an angle of \(55^{\circ}\) below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?

Short Answer

Expert verified
The horizontal component of the ball's velocity is approximately 8.6 m/s.

Step by step solution

01

Understand the Problem

We need to find the horizontal component of the volleyball's velocity. The ball is spiked at a speed of \(15 \, \text{m/s}\) at a \(55^\circ\) angle below the horizontal.
02

Component Formulas

To find the horizontal component (\(v_x\)) of the velocity when the ball is hit at an angle (\(\theta\)), we use the equation: \(v_x = v \cdot \cos(\theta)\), where \(v\) is the initial velocity of \(15 \, \text{m/s}\) and \(\theta = 55^\circ\).
03

Calculate the Horizontal Component

Substitute the given values into the formula: \[ v_x = 15 \cdot \cos(55^\circ) \]
04

Use Trigonometric Values

Calculate \(\cos(55^\circ)\) using a calculator: \(\cos(55^\circ) \approx 0.5736\).
05

Final Calculation

Complete the calculation: \[ v_x = 15 \cdot 0.5736 = 8.604 \, \text{m/s} \]

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Key Concepts

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

Horizontal Component
In the realm of projectile motion, the horizontal component of velocity plays a vital role. It represents the portion of an object's velocity that moves parallel to the ground. When a volleyball is spiked, it needs to be broken down into horizontal and vertical movements to fully comprehend its path.
  • The horizontal component remains constant if air resistance is negligible.
  • It influences how far and how quickly the projectile moves horizontally.
To calculate the horizontal component from an initial velocity, we use the formula:
\[ v_x = v \cdot \cos(\theta) \] where \(v_x\) is the horizontal velocity, \(v\) is the initial velocity, and \(\theta\) is the angle of projection. Understanding this concept helps predict where and when the ball will reach its destination.
Initial Velocity
Initial velocity is a key concept in understanding projectile motion. It is the velocity with which an object is launched. In our volleyball scenario, the initial velocity is given as 15 m/s. This velocity is not directed purely horizontal or vertical but at an angle beneath the horizontal.
  • It dictates the starting speed and direction of the object.
  • Components of initial velocity determine the object's future motion.
Decomposing this initial velocity into horizontal and vertical components allows us to analyze and predict the object's trajectory accurately. It's crucial to remember that the initial speed and angle directly affect both the range and the height the projectile will achieve.
Trigonometric Functions
Trigonometric functions are mathematical tools that help decompose initial velocity into components. In projectile motion, these functions tell us what part of the velocity acts in the horizontal and vertical directions.
  • They make use of angles to describe these components.
  • Cosine (\(\cos\)) is used for horizontal components.
  • Sine (\(\sin\)) is used for vertical components.
For instance, in our example, we utilize \(\cos(\theta)\) to find the horizontal component: \[ v_x = v \cdot \cos(\theta) \] Knowing these trigonometric principles is essential for solving projectile problems, as they bridge the initial conditions to real-world motion outcomes.

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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?

Useful background for this problem can be found in Multiple-Concept Example 2. On a spacecraft two engines fire for a time of 565 s. One gives the craft an acceleration in the \(x\) direction of \(a_{x}=5.10 \mathrm{m} / \mathrm{s}^{2},\) while the other produces an acceleration in the \(y\) direction of \(a_{y}=7.30 \mathrm{m} / \mathrm{s}^{2} .\) At the end of the firing period, the craft has velocity components of \(v_{x}=3775 \mathrm{m} / \mathrm{s}\) and \(v_{y}=4816 \mathrm{m} / \mathrm{s} .\) Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the \(+x\) axis.

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?

A skateboarder, starting from rest, rolls down a 12.0 -m ramp. When she arrives at the bottom of the ramp her speed is \(7.70 \mathrm{m} / \mathrm{s}\). (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at \(25.0^{\circ}\) with respect to the ground, what is the component of her acceleration that is parallel to the ground?

The highest barrier that a projectile can clear is \(13.5 \mathrm{m},\) when the projectile is launched at an angle of \(15.0^{\circ}\) above the horizontal. What is the projectile's launch speed?
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