Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 64GP
To determine

To calculate: Minimum speed required by the ball to clear the net, distance at which the ball lands and the time for which the ball will be in the air. Also state if the ball lands at a good distance.

Expert Solution & Answer
Check Mark

Answer to Problem 64GP

Minimum speed required by the ball to clear the net is 2.63 m/s.

Distance at which the ball lands is 18.7 m. Since, this is within 15 m to 22 m, the ball is in good region.

Time for which the ball is in the air is 0.71 s.

Explanation of Solution

Given:

Height of the net = 0.90 m

Distance between net and the server = 15 m

Height at which the ball is launched = 2.50 m

Formula used:

The following kinematic equations:

  y=y0+vy0t+12ayt2

Where, y is displacement, y0 is the initial position, vy0 is the initial velocity in the vertical direction, t is time taken and ay ( 9.8 m/s2 ) is acceleration.

  Δx=Vxt

Where, Δx is the horizontal distance, Vx is velocity in the horizontal direction and t is time taken.

Calculation:

Initial position of the ball is taken as 2.50 m, initial velocity is taken as 0, acceleration is taken as negative (against gravity) and y is taken as height of the net.

Time taken for the ball to reach the net is calculated as:

  y=y0+vy0t+12ayt20.90=2.50+0t+12(9.8)t2 0.9=2.504.9t24.9t2=1.6t2=1.64.9t=1.64.9t=0.57 s

Therefore, time for which the ball to reach the net 0.57 seconds.

Distance between net and server is taken as horizontal distance.

Velocity is calculated as:

  Δx=VxtVx=ΔxtVx=15.00.57Vx=26.3 m/s

Therefore, minimum required to clear the net is 2.63 m/s.

To find time for which the ball is in the air, the final displacement, y should be 0. Therefore,

  y=y0+vy0t+12ayt20=2.50+0t+12(9.8)t24.9t2=2.50t2=2.504.9t=2.504.9t=0.71 s

Therefore, time for which the ball is in the air is 0.71 seconds.

Distance at which the ball lands is calculated as:

  Δx=VxtΔx=(26.3)(0.71)Δx=18.7 m

Therefore, distance at which the ball lands is 18.7 m.

Since, this is within 15 m to 22 m, the ball is in good region.

Conclusion:

Minimum speed required by the ball to clear the net is 2.63 m/s.

Distance at which the ball lands is 18.7 m. Since, this is within 15 m to 22 m, the ball is in good region.

Time for which the ball is in the air is 0.71 s.

Chapter 3 Solutions

Physics: Principles with Applications

Ch. 3 - Prob. 11QCh. 3 - Prob. 12QCh. 3 - Prob. 13QCh. 3 - A projectile is launched at an upward angle of 300...Ch. 3 - Prob. 15QCh. 3 - Two cannonballs, A and B, are fired from the...Ch. 3 - 18. A person sitting in an enclosed train car,...Ch. 3 - Prob. 18QCh. 3 - Prob. 19QCh. 3 - Prob. 20QCh. 3 - A car is driven 225 km west and then 98 km...Ch. 3 - A delivery truck travels 21 blocks north, 16...Ch. 3 - If Vx=9.80 units and Vy=6.40 units, determine the...Ch. 3 - Graphically determine the resultant of the...Ch. 3 - V is a vector 24.8 units in magnitude and points...Ch. 3 - Vector V is 6.6 using long and points along the...Ch. 3 - Figure 3-33 shows two vectors, A and B , whose...Ch. 3 - Prob. 8PCh. 3 - Three vectors are shown in Fig. 3-35 Q. Their...Ch. 3 - (a) given the vectors A and B shown in Fig. 3-35,...Ch. 3 - Determine the vector AC , given the vectors A and...Ch. 3 - For the vectors shown in Fig. 3—35, determine (a)...Ch. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - 17. (l) A tiger leaps horizontally from a...Ch. 3 - 18. (l) A diver running 2.5 m/s dives out...Ch. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53GPCh. 3 - Prob. 54GPCh. 3 - Prob. 55GPCh. 3 - Prob. 56GPCh. 3 - Prob. 57GPCh. 3 - Prob. 58GPCh. 3 - Prob. 59GPCh. 3 - Prob. 60GPCh. 3 - Prob. 61GPCh. 3 - Prob. 62GPCh. 3 - Prob. 63GPCh. 3 - Prob. 64GPCh. 3 - Prob. 65GPCh. 3 - Prob. 66GPCh. 3 - Prob. 67GPCh. 3 - Prob. 68GPCh. 3 - Prob. 69GPCh. 3 - Prob. 70GPCh. 3 - Prob. 71GPCh. 3 - Prob. 72GPCh. 3 - Prob. 73GPCh. 3 - Prob. 74GPCh. 3 - Prob. 75GP

Additional Science Textbook Solutions

Find more solutions based on key concepts
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
Kinematics Part 3: Projectile Motion; Author: Professor Dave explains;https://www.youtube.com/watch?v=aY8z2qO44WA;License: Standard YouTube License, CC-BY