Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Videos

Question
Book Icon
Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution
Check Mark

Answer to Problem 94P

The distance of the ball strikes the wall is 1.1m.

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

Δx=vxΔt (I)

Here, Δx is the displacement of the ball, vx is the velocity of the ball along x axis, and Δt is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

vx=vicosθ (II)

Here, vi is the initial velocity of the ball and θ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

viy=visinθ (III)

Here, vi is the initial velocity of the ball and θ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

Δy=viyΔt+12ayΔt2

Here, Δy is the change in height of the ball, viy is the initial velocity of the ball along y axis, and ay is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

Δy=viyΔt12gΔt2 (IV)

Here, g is the gravitational acceleration.

Write the equation for change in height of the ball.

Δy=yfyi (V)

Here, yf is the final height and yi is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find Δt.

Δx=(vicosθ)ΔtΔt=Δxvicosθ (VI)

Substitute equation (III) and (VI) in equation (IV) to find Δy.

Δy=(visinθ)Δt12gΔt2=(visinθ)Δxvicosθ12g(Δx2vi2cos2θ)=Δxtanθ(gΔx22vi2cos2θ)

Substitute the above relation in equation (V) to find yf.

yf=yi+Δy=yi+Δxtanθ(gΔx22vi2cos2θ)

Substitute 60cm for yi, 10m for Δx, 9.80m/s2 for g, 20m/s for vi, and 90°80°=10° for θ in the above equation.

yf=60cm(0.01m1cm)+(10m)tan10°((9.80m/s2)(10m)22(20m/s)2cos210°)=2.36m980m3/s2775.9m2/s2=2.36m1.26m=1.1m

Therefore, the distance of the ball strikes the wall is 1.1m.

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

(b)

Expert Solution
Check Mark

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

Δv=vfyviy (VII)

Here, vfy is the final velocity and viy is the initial velocity.

Write the equation of motion for the acceleration of ball.

Δv=ayΔt (VIII)

Here, Δv is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

Δv=gΔt (IX)

Conclusion:

Compare equation (VII) and (IX).

vfyviy=gΔt

Replace visinθ for viy and Δxvicosθ for Δt in above relation.

vfyvisinθ=g(Δxvicosθ)vfy=visinθ(gΔxvicosθ) (X)

Substitute 10m for Δx, 9.80m/s2 for g, 20m/s for vi, and 90°80°=10° for θ in equation (X).

vfy=(20m/s)sin10°((9.80m/s2)(10m)(20m/s)cos10°)=3.47m/s4.98m/s=1.51m/s1.5m/s

Since,vfy<0, so that ball is on its way down.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Given two particles with Q = 4.40-µC charges as shown in the figure below and a particle with charge q = 1.40 ✕ 10−18 C at the origin. (Note: Assume a reference level of potential V = 0 at r = ∞.) Three positively charged particles lie along the x-axis of the x y coordinate plane.Charge q is at the origin.Charge Q is at (0.800 m, 0).Another charge Q is at (−0.800 m, 0).(a)What is the net force (in N) exerted by the two 4.40-µC charges on the charge q? (Enter the magnitude.) N(b)What is the electric field (in N/C) at the origin due to the two 4.40-µC particles? (Enter the magnitude.) N/C(c)What is the electrical potential (in kV) at the origin due to the two 4.40-µC particles? kV(d)What If? What would be the change in electric potential energy (in J) of the system if the charge q were moved a distance d = 0.400 m closer to either of the 4.40-µC particles?
(a) Where does an object need to be placed relative to a microscope in cm from the objective lens for its 0.500 cm focal length objective to produce a magnification of -25? (Give your answer to at least three decimal places.) 0.42 × cm (b) Where should the 5.00 cm focal length eyepiece be placed in cm behind the objective lens to produce a further fourfold (4.00) magnification? 15 × cm
In a LASIK vision correction, the power of a patient's eye is increased by 3.10 D. Assuming this produces normal close vision, what was the patient's near point in m before the procedure? (The power for normal close vision is 54.0 D, and the lens-to-retina distance is 2.00 cm.) 0.98 x m

Chapter 3 Solutions

Physics

Ch. 3.5 - Prob. 3.6PPCh. 3.5 - Prob. 3.5ACPCh. 3.5 - Prob. 3.7PPCh. 3.5 - Prob. 3.5BCPCh. 3.6 - Prob. 3.6CPCh. 3.6 - Prob. 3.8PPCh. 3.6 - Prob. 3.9PPCh. 3 - Prob. 1CQCh. 3 - Prob. 2CQCh. 3 - Prob. 3CQCh. 3 - Prob. 4CQCh. 3 - Prob. 5CQCh. 3 - Prob. 6CQCh. 3 - Prob. 7CQCh. 3 - Prob. 8CQCh. 3 - Prob. 9CQCh. 3 - Prob. 10CQCh. 3 - Prob. 11CQCh. 3 - Prob. 12CQCh. 3 - Prob. 13CQCh. 3 - Prob. 14CQCh. 3 - Prob. 15CQCh. 3 - Prob. 1MCQCh. 3 - Prob. 2MCQCh. 3 - 4. A runner moves along a circular track at a...Ch. 3 - Prob. 4MCQCh. 3 - Prob. 5MCQCh. 3 - Prob. 6MCQCh. 3 - Prob. 7MCQCh. 3 - Prob. 8MCQCh. 3 - Prob. 9MCQCh. 3 - Prob. 10MCQCh. 3 - Prob. 11MCQCh. 3 - Prob. 12MCQCh. 3 - Prob. 13MCQCh. 3 - Prob. 14MCQCh. 3 - Prob. 15MCQCh. 3 - Prob. 16MCQCh. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - 12. Michaela is planning a trip in Ireland from...Ch. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 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. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 60PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 68PCh. 3 - Prob. 69PCh. 3 - Prob. 70PCh. 3 - Prob. 71PCh. 3 - Prob. 72PCh. 3 - 75. A Nile cruise ship takes 20.8 h to go upstream...Ch. 3 - Prob. 74PCh. 3 - Prob. 75PCh. 3 - Prob. 76PCh. 3 - Prob. 77PCh. 3 - Prob. 78PCh. 3 - Prob. 79PCh. 3 - Prob. 80PCh. 3 - Prob. 81PCh. 3 - Prob. 82PCh. 3 - Prob. 83PCh. 3 - Prob. 84PCh. 3 - Prob. 85PCh. 3 - Prob. 86PCh. 3 - Prob. 87PCh. 3 - Prob. 88PCh. 3 - Prob. 89PCh. 3 - Prob. 90PCh. 3 - Prob. 91PCh. 3 - Prob. 92PCh. 3 - Prob. 93PCh. 3 - Prob. 94PCh. 3 - Prob. 96PCh. 3 - Prob. 95PCh. 3 - Prob. 97PCh. 3 - Prob. 98PCh. 3 - Prob. 99PCh. 3 - Prob. 100PCh. 3 - Prob. 101PCh. 3 - Prob. 102PCh. 3 - Prob. 103PCh. 3 - Prob. 104PCh. 3 - Prob. 105PCh. 3 - Prob. 106PCh. 3 - Prob. 107PCh. 3 - Prob. 108PCh. 3 - 111. A ball is thrown horizontally off the edge of...Ch. 3 - 112. A marble is rolled so that it is projected...Ch. 3 - Prob. 111PCh. 3 - Prob. 112PCh. 3 - Prob. 113PCh. 3 - Prob. 114P
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
Vectors and 2D Motion: Crash Course Physics #4; Author: CrashCourse;https://www.youtube.com/watch?v=w3BhzYI6zXU;License: Standard YouTube License, CC-BY