bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 20P

(a)

To determine

The maximum height reached by the rock.

(a)

Expert Solution
Check Mark

Answer to Problem 20P

The maximum height reached by the rock is 77.051 m

Explanation of Solution

Write the equation of motion.

  v2u2=2as        (I)

Here, v is the final velocity, u is the initial velocity, a is the acceleration, and s is the height.

Conclusion:

Substitute 0 for v, 0.6342 for u, -9.8m/s2 for a in expression (I)

    v2u2=2as06.342=2×(9.8)×ss=2.051m

Therefore the maximum height of the rock reach is 2.051m plus the height of the bridge,

  75+2.051 = 77.051 m

(b)

To determine

The time required by the rock to reach the maximum height.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

The time required by the rock to reach the maximum height is 0.647sec

Explanation of Solution

Write the equation of motion.

  v=u+at        (I)

Here, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval.

Conclusion:

Substitute 0 for v, 0.6342 for u, -9.8m/s2 for a in expression (I)

  0 = 6.349.8 tt = 0.647sec

The time required by the rock to reach the maximum height is 0.647sec

(c)

To determine

The place where the rock lands.

(c)

Expert Solution
Check Mark

Answer to Problem 20P

The place where the rock lands is 62.7m

Explanation of Solution

Write the equation of motion.

  s=ut+12at2        (I)

Here, u is the initial velocity, a is the acceleration, s is the height, and t is the time interval.

Conclusion:

Substitute 0 for u, 77.051 for u, -9.8m/s2 for a in expression (I)

  s=ut+12at277.051=0+0.5×9.8ktt2t=3.965s

Total time taken to reach ground is 3.965 +0.647 = 4.612 s

Therefore the total horizontal displacement is,

  xlands =v0xtlands         (II)

Substitute 15m/s for v0x, 4.61s for tlands in expression (II)

  xlands=(15m/s)(cos25°)(4.61s)=63 m

The place where the rock lands is 62.7m

(d)

To determine

The time at which the rock lands.

(d)

Expert Solution
Check Mark

Answer to Problem 20P

The time at which the rock lands is  4.612 s

Explanation of Solution

Write the equation of motion.

  s=ut+12at2        (I)

Here, u is the initial velocity, a is the acceleration, s is the height, and t is the time interval.

Conclusion:

Substitute 0 for u, 77.051 for u, -9.8m/s2 for a in expression (I)

  s=ut+12at277.051=0+0.5×9.8ktt2t=3.965s

Total time taken to reach ground is 3.965 +0.647 = 4.612 s

The time at which the rock lands is  4.612 s

(e)

To determine

The velocity of the rock just before it lands.

(e)

Expert Solution
Check Mark

Answer to Problem 20P

The magnitude of the velocity 41.167 m/s and 70.72°.

Explanation of Solution

Write the equation of motion.

  v=u+at        (I)

Here, u is the initial velocity, a is the acceleration, s is the height, and t is the time interval.

Conclusion:

Substitute 0 for u, 3.965 for t, 9.8m/s2 for a in expression (I)

  vy=u+atvy= 0 +9.8×3.965=38.857 m/s

And the velocity along vx= 13.595 m/s

The magnitude of the velocity,

  |v|= (13.595 m/s)2+(38.857)2= 41.167 m/s

Write the direction will then be,

   θ=tan1(vyvx) θ=tan1(38.9 m/s13.6 m/s)θ=70.72°

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 4 Solutions

Bundle: College Physics: Reasoning And Relationships, 2nd + Webassign Printed Access Card For Giordano's College Physics, Volume 1, 2nd Edition, Multi-term

Ch. 4 - Prob. 5QCh. 4 - Prob. 6QCh. 4 - Prob. 7QCh. 4 - Prob. 8QCh. 4 - Prob. 9QCh. 4 - Prob. 10QCh. 4 - Prob. 11QCh. 4 - Prob. 12QCh. 4 - Prob. 13QCh. 4 - Prob. 14QCh. 4 - Prob. 15QCh. 4 - Prob. 16QCh. 4 - Prob. 17QCh. 4 - Prob. 18QCh. 4 - Prob. 19QCh. 4 - Prob. 20QCh. 4 - Prob. 1PCh. 4 - Prob. 2PCh. 4 - Several forces act on a particle as shown in...Ch. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - The sled in Figure 4.2 is stuck in the snow. A...Ch. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - A bullet is fired from a rifle with speed v0 at an...Ch. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - An airplane flies from Boston to San Francisco (a...Ch. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Two crates of mass m1 = 35 kg and m2 = 15 kg are...Ch. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Consider the motion of a bicycle with air drag...Ch. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 70PCh. 4 - Prob. 71PCh. 4 - Prob. 72PCh. 4 - Prob. 73PCh. 4 - Prob. 74PCh. 4 - A vintage sports car accelerates down a slope of ...Ch. 4 - Prob. 76PCh. 4 - Prob. 77PCh. 4 - Prob. 78PCh. 4 - Prob. 79PCh. 4 - Prob. 80PCh. 4 - Prob. 81PCh. 4 - Prob. 82PCh. 4 - Prob. 83PCh. 4 - Prob. 84PCh. 4 - Prob. 85PCh. 4 - Prob. 86PCh. 4 - Two blocks of mass m1 = 2.5 kg and m2 = 3.5 kg...Ch. 4 - Prob. 88PCh. 4 - Prob. 89PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92P
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
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: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
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