/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 96 For women's volleyball the top o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For women's volleyball the top of the net is \(2.24 \mathrm{~m}\) above the floor and the court measures \(9.0 \mathrm{~m}\) by \(9.0 \mathrm{~m}\) on each side of the net. Using a jump serve, a player strikes the ball at a point that is \(3.0 \mathrm{~m}\) above the floor and a horizontal distance of \(8.0 \mathrm{~m}\) from the net. If the initial velocity of the ball is horizontal, (a) what minimum magnitude must it have if the ball is to clear the net and (b) what maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net?

Short Answer

Expert verified
To clear the net, the minimum speed is about 6.92 m/s, and to land within bounds, the maximum speed is about 10.23 m/s.

Step by step solution

01

Understand the Problem

The ball is hit from a height of 3.0 m and must clear a 2.24 m high net 8.0 m away. After passing the net, it should land within 9.0 m on the opposite side. The initial velocity is only horizontal; we need to find its minimum and maximum magnitudes to meet these criteria.
02

Use Projectile Motion Equations

Write the equations for vertical motion under gravity: the height of the ball as a function of time is given by \( y = y_0 + v_{y0}t - \frac{1}{2}gt^2 \), where \( y_0 = 3.0 \text{ m} \), \( v_{y0} = 0 \), and \( g = 9.81 \text{ m/s}^2 \). The horizontal motion and distance covered is \( x = v_{x}t \), where \( v_x \) is the horizontal velocity.
03

Determine Time to Clear the Net

Find the time \( t \) it takes for the ball to clear the net. The vertical position at the net (2.24 m) should satisfy \( 2.24 = 3.0 - \frac{1}{2}(9.81)t^2 \). Solve for \( t \) to find when it clears the net.
04

Solve for Time \( t \) at the Net Height

Solve \( 3.0 - \frac{1}{2}(9.81)t^2 = 2.24 \). Simplifying, we get \( \frac{1}{2}g t^2 = 0.76 \). Solve this for \( t \) using \( t = \sqrt{\frac{2 \times 0.76}{g}} \).
05

Calculate Minimum Initial Velocity

Substitute back into the horizontal motion equation \( 8.0 = v_x t \) using the calculated \( t \) to find \( v_x \), i.e., \( v_x = \frac{8.0}{t} \). This \( v_x \) is the minimum initial velocity to clear the net.
06

Calculate Maximum Initial Velocity

To remain inbounds on the opposite side, the total horizontal distance is 17.0 m (8 m to the net plus 9 m on the opposite court). The time \( T \) to hit the floor is given by \( 3 = \frac{1}{2}gT^2 \). Substituting into the horizontal equation gives \( v_x \leq \frac{17}{T} \). Solve \( T = \sqrt{\frac{2 \times 3}{g}} \) and substitute to find the maximum velocity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Horizontal Velocity
In projectile motion, horizontal velocity refers to the component of the velocity that moves the object along the horizontal plane. For a volleyball hit with a jump serve, the initial velocity is entirely horizontal, meaning it does not initially contribute to the ball's height. This horizontal velocity remains constant throughout the flight, as no horizontal forces act on the ball (assuming air resistance is negligible).
Understanding horizontal velocity is crucial to ensure that the ball covers the necessary horizontal distances:
  • At least 8 meters to clear the net.
  • No more than 17 meters to stay in bounds, as it must not exceed the total court length of 17 meters (8 meters to the net and 9 meters beyond).
By managing horizontal velocity correctly, one can calculate the speed needed to clear obstacles like the volleyball net and land within boundaries. This calculation involves determining the time it takes for the ball to cover specific horizontal distances and then using these times to find the minimum and maximum permissible horizontal velocities.
Vertical Motion
Vertical motion in projectile physics accounts for the object's movement against gravity. For the volleyball jump serve, while the initial vertical velocity is zero (since all the serve's speed is horizontal), gravity influences the upward and downward motion throughout the flight.
The formula to describe vertical motion is given by \[ y = y_0 + v_{y0}t - \frac{1}{2}gt^2 \]where:
  • \( y_0 \) is the initial height above the floor (3.0 m).
  • \( g \) is the acceleration due to gravity, \( 9.81 \text{ m/s}^2 \).
  • \( t \) is the time of flight.
Gravity acts to slow the ball's ascent and increase its descent, making it crucial to calculate the time this motion takes to clear the net at 2.24 m. Using the above formula helps determine crucial time values needed for both minimum and maximum initial velocity calculations.
Gravity
Gravity plays a vital role in determining the trajectory of any projectile, including a volleyball in play. It is the force that pulls the ball back toward the Earth, affecting its vertical motion.
In our context, \( g = 9.81 \text{ m/s}^2 \) represents the acceleration due to gravity. From the moment the ball is struck, gravity causes it to follow a parabolic path:
  • Initially, gravity reduces the ball's vertical height as it approaches the net.
  • Then, after reaching its highest point, gravity accelerates the ball downward until it lands.
Gravity must be accounted for when calculating how long it takes for the ball to travel vertical distances. Solving vertical motion equations with gravity is how we find the exact times involved for net clearance and landing on the opponent's court.
Volleyball Physics
Volleyball physics involves principles of both horizontal velocity and vertical motion under gravity. In a volleyball jump serve, mastering these elements ensures the serve is both strategic and effective.
To break it down:
  • A well-calculated serve ensures the ball crosses above the net with enough speed, optimizing horizontal velocity so the game is in your favor.
  • Understanding vertical motion helps players judge how the ball's height changes with time as it approaches the net, combining calculations for how gravity affects it.
  • Choosing the right velocities means the ball lands in the designated area, making it challenging for opponents to return.
In competitive volleyball, physics isn't just a theory. It's applied practically for plays that are precise and consistent. Players and coaches often rely on these physical principles to refine their techniques and outsmart their opponents. Proper understanding of these mechanisms enhances performance, both during practice and in actual games.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The position \(\vec{r}\) of a particle moving in an \(x y\) plane is given by \(\vec{r}=\left(2.00 t^{3}-5.00 t\right) \hat{\mathrm{i}}+\left(6.00-7.00 t^{4}\right) \hat{\mathrm{j}}\), with \(\vec{r}\) in meters and \(\mathrm{t}\) in seconds. In unit-vector notation, calculate (a) \(\vec{r},(\mathrm{~b}) \vec{v}\), and \((\mathrm{c}) \vec{a}\) for \(t=2.00 \mathrm{~s}\) (d) What is the angle between the positive direction of the \(x\) axis and a line tangent to the particle's path at \(t=2.00 \mathrm{~s}\) ?

A football player punts the football so that it will have a "hang time" (time of flight) of \(4.5 \mathrm{~s}\) and land \(46 \mathrm{~m}\) away. If the ball leaves the player's foot \(150 \mathrm{~cm}\) above the ground, what must be the (a) magnitude and (b) angle (relative to the horizontal) of the ball's initial velocity?

An electron having an initial horizontal velocity of magnitude \(1.00 \times 10^{9} \mathrm{~cm} / \mathrm{s}\) travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of \(2.00 \mathrm{~cm}\) and has a constant downward acceleration of magnitude \(1.00 \times 10^{17} \mathrm{~cm} / \mathrm{s}^{2}\) due to the charged plates. Find (a) the time the electron takes to travel the \(2.00 \mathrm{~cm},(\mathrm{~b})\) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

A soccer ball is kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ} .\) A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?

A light plane attains an airspeed of \(500 \mathrm{~km} / \mathrm{h}\). The pilot sets out for a destination \(800 \mathrm{~km}\) due north but discovers that the plane must be headed \(20.0^{\circ}\) east of due north to fly there directly. The plane arrives in \(2.00 \mathrm{~h}\). What were the (a) magnitude and (b) direction of the wind velocity?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.