Homework Help is Here – Start Your Trial Now!
Science
Physics
I just need the practice problem solved with work shown. The rest of the images provides the information needed to solve the problem. The correct answer is also provided to check work

I just need the practice problem solved with work shown. The rest of the images provides the information needed to solve the problem. The correct answer is also provided to check work

College Physics
College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter2: Kinematics
Section: Chapter Questions
Problem 33PE: An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s...
See similar textbooks
icon
Related questions
Question

I just need the practice problem solved with work shown. The rest of the images provides the information needed to solve the problem. The correct answer is also provided to check work. 

EXAMPLE 2.10 Falling euro in Pisa Our first example of free-fall motion is very simple. A 2-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest and falls freely. Ignore the effects of air resistance, and compute its position and velocity after 1.00 s, 2.00 s, and 3.00 s.
Transcribed Image Text:EXAMPLE 2.10 Falling euro in Pisa Our first example of free-fall motion is very simple. A 2-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest and falls freely. Ignore the effects of air resistance, and compute its position and velocity after 1.00 s, 2.00 s, and 3.00 s.
SOLUTION SET UP Figure 2.25 shows the diagram we sketch. For a coordinate system, we choose a vertical axis with the upward direction to be posi- tive, and we place the origin at the starting point. The position, velocity, and acceleration have only y components. The initial value yo of the co- ordinate y and the initial y component of velocity voy are both zero. The acceleration is downward (in the negative y direction), so the constant y component of acceleration is ay = -g = -9.80 m/s². (Remember that, by definition, g itself is always positive.) SOLVE From Equations 2.13 and 2.12 we have y = vot + a₂t² = 0 + (−8)t² = (-4.90 m/s²)t², SW BOULU05 DIN 100 mort, + ayt = 0 + (−g)t = (-9.80 m/s²)t. Vy = Voy og Amoer - Yel (alm 2 s, y₁ Using subscript 1 for the first time, we find that when t = t₁ = 1.00 1.00 s, y₁ = (-4.90 m/s²) (1.00 s)² = -4.90 m and v₁y = Uly (-9.80 m/s2)(1.00 s) = -9.80 m/s. The coin is therefore 4.90 m be- low the origin (because y is negative) and has a downward velocity (uy is negative) with magnitude 9.80 m/s. ob The positions and velocities at 2.00 s and 3.00 s are found in the same way. The results are -19.6 m and -19.6 m/s at t = 2.00 s and -44.1 m and -29.4 m/s at t = 3.00 s. REFLECT All our values of y and v, are negative because we have cho- sen the positive y axis to be upward. We could just as well have chosen the downward direction to be positive. In that case, the y component of acceleration would have been a, = g = +9.80 m/s². The crucial point +9.80 m/s². The crucial that you must decide at the start which direction is positive, and then stick with your decision. mioq dgu point vizu odit soulsy ovings zli tud Jmclzni as ch at 0195 gr play The Leaning Tower *** sty din 2.6 Freely Falling Objects 49 CHON ill LATION B 1435 al FIGURE 2.25 Our sketch for the problem eft im Y = 0 Y 0 V₁y = ? Vzy=? V3Y =? to = 0, yo=0 t₁ = 100 s, y₁= ? ay=-9--980 m/s² t₂ = 2.00 s₂ Y₂ = ? - 13=3.00 S, y3 = ? Practice Problem: The coin is released 47.0 m above the ground. How much time is required for the coin to fall to the ground? What is its velocity just before it strikes the ground? Answers: 3.10 s, -30.4 m/s. how of snuf ow ob anoires 18// 39303
Transcribed Image Text:SOLUTION SET UP Figure 2.25 shows the diagram we sketch. For a coordinate system, we choose a vertical axis with the upward direction to be posi- tive, and we place the origin at the starting point. The position, velocity, and acceleration have only y components. The initial value yo of the co- ordinate y and the initial y component of velocity voy are both zero. The acceleration is downward (in the negative y direction), so the constant y component of acceleration is ay = -g = -9.80 m/s². (Remember that, by definition, g itself is always positive.) SOLVE From Equations 2.13 and 2.12 we have y = vot + a₂t² = 0 + (−8)t² = (-4.90 m/s²)t², SW BOULU05 DIN 100 mort, + ayt = 0 + (−g)t = (-9.80 m/s²)t. Vy = Voy og Amoer - Yel (alm 2 s, y₁ Using subscript 1 for the first time, we find that when t = t₁ = 1.00 1.00 s, y₁ = (-4.90 m/s²) (1.00 s)² = -4.90 m and v₁y = Uly (-9.80 m/s2)(1.00 s) = -9.80 m/s. The coin is therefore 4.90 m be- low the origin (because y is negative) and has a downward velocity (uy is negative) with magnitude 9.80 m/s. ob The positions and velocities at 2.00 s and 3.00 s are found in the same way. The results are -19.6 m and -19.6 m/s at t = 2.00 s and -44.1 m and -29.4 m/s at t = 3.00 s. REFLECT All our values of y and v, are negative because we have cho- sen the positive y axis to be upward. We could just as well have chosen the downward direction to be positive. In that case, the y component of acceleration would have been a, = g = +9.80 m/s². The crucial point +9.80 m/s². The crucial that you must decide at the start which direction is positive, and then stick with your decision. mioq dgu point vizu odit soulsy ovings zli tud Jmclzni as ch at 0195 gr play The Leaning Tower *** sty din 2.6 Freely Falling Objects 49 CHON ill LATION B 1435 al FIGURE 2.25 Our sketch for the problem eft im Y = 0 Y 0 V₁y = ? Vzy=? V3Y =? to = 0, yo=0 t₁ = 100 s, y₁= ? ay=-9--980 m/s² t₂ = 2.00 s₂ Y₂ = ? - 13=3.00 S, y3 = ? Practice Problem: The coin is released 47.0 m above the ground. How much time is required for the coin to fall to the ground? What is its velocity just before it strikes the ground? Answers: 3.10 s, -30.4 m/s. how of snuf ow ob anoires 18// 39303
Expert Solution
Step 1

Physics homework question answer, step 1, image 1

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Power
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
College Physics
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College
Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers with Modern …
Physics for Scientists and Engineers with Modern …
Physics
ISBN:
9781337553292
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers: Foundations…
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS