Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 102P

(a)

To determine

The value of angle formed by initial velocity vector with horizontal.

(a)

Expert Solution
Check Mark

Answer to Problem 102P

  θo=86.38°

Explanation of Solution

Given data:

Final height of ball just above the plate =0.7 m

Range =18.4 m

Initial speed of ball =37.5 m/s

Formula used:

Vertical position of ball in terms of time t .

  y=yo+(vosinθo)t12gt2

Calculation:

The horizontal distance covered is given as

  x=(vocosθo)t                      ...(1)

The vertical position can be written as

  y=yo+(vosinθo)t12gt2

From equation (1), put t=xvocosθo

  y=yo+(vosinθo)t12gt2y=yo+(vosinθo)xvocosθo12g( x v o cos θ o )2y=yo(tanθo)xg2vo2 cos2θox2                      ...(2)

Now, substitute the values of respective parameters in equation

  y=yo(tanθo)xg2vo2 cos2θox20.7=2.0(tanθo)×18.49.812× ( 37.5 )2× cos2θo(18.4)218.4×tanθo+1.181 cos2θo=1.318.4×tanθo+1.181 cos2θo=1.318.4× sec2θo1+1.181 cos2θo=1.318.4× 1 cos 2 θ o1+1.181 cos2θo=1.318.4× 1 cos 2 θ o1=1.31.181 cos2θo

By squaring both sides,

  (18.4× 1 cos 2 θ o 1)2=(1.3 1.181 cos 2 θ o )2338.56 cos 2θo338.56=1.69+1.394 cos4θo3.07 cos2θo1.394 cos4θo341.63 cos2θo+340.25=0

After solving the above equation, the value of θo

  1.394 cos4θo341.63 cos2θo+340.25=01 cos2θo=( 341.63)± ( 341.63 ) 2 4×1.394×340.252×1.3941 cos2θo=341.63± 116711.05691897.2342.71 cos2θo=341.63±338.842.7

By taking the positive sign

  1 cos2θo=341.63+338.842.71 cos2θo=252.025cos2θo=1252.025cos2θo=3.967×103cosθo=3.967× 10 3cosθo=0.06299θo=cos1(0.06299)θo=86.38°

Conclusion:

The velocity and the horizon made an angle is θo=86.38°

(b)

To determine

The value of speed with which ball crosses plate.

(b)

Expert Solution
Check Mark

Answer to Problem 102P

  v=38.88 m/s

Explanation of Solution

Given data:

Final height of ball just above the plate =0.7 m

Range =18.4 m

Initial speed of ball =37.5 m/s

Formula used:

Speed is given by

  v=vx2+vy2

Calculation:

The horizontal component of velocity is vx

  vx=dxdt=ddt(vocosθo)t=vocosθo

The vertical component of velocity is vy

  vy=dydt=ddt(yo+vosinθo×t12gt2)=vosinθogt

Substitute these values into speed formula

  v= ( v o cos θ o )2+ ( v o sin θ o gt )2v= ( v o cos θ o )2+ ( v o sin θ o g x v o cos θ o )2v= ( 37.5×cos86.38° )2+ ( 37.5×sin86.38°9.81× 18.4 37.5×cos86.38° )2v=5.606+1506.2566v=38.88 m/s

Conclusion:

The velocity of the ball over the plate is 38.88 m/s .

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
Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC
In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?
Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor  µC 6.00 µF capacitor  µC 3.00 µF capacitor  µC capacitor C  µC

Chapter 3 Solutions

Physics for Scientists and Engineers

Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 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 - Prob. 73PCh. 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. 95PCh. 3 - Prob. 96PCh. 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 - Prob. 109PCh. 3 - Prob. 110PCh. 3 - Prob. 111PCh. 3 - Prob. 112PCh. 3 - Prob. 113PCh. 3 - Prob. 114PCh. 3 - Prob. 115PCh. 3 - Prob. 116PCh. 3 - Prob. 117PCh. 3 - Prob. 118PCh. 3 - Prob. 119PCh. 3 - Prob. 120PCh. 3 - Prob. 121PCh. 3 - Prob. 122P
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
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Kinematics Part 3: Projectile Motion; Author: Professor Dave explains;https://www.youtube.com/watch?v=aY8z2qO44WA;License: Standard YouTube License, CC-BY