Good day to youI was looking through a particular problem found in Richard Wolfson Essential University Physics Volume one, 4th Edition textbook, on chapter 3, problem 68. Your solution derives an inital speed equation, however, I don't know how it was derived. was it a manipulation of the x and y components of the v2 = v0^2 +2as or something else entirely?Thanks in advance < Chapter 3, Problem 68PThe Gap between the buildings is 4.5 m wide.The stuntman runs from one building roof to another buildingroof. The height between two building roofs is 1.9 m.Calculation:At the beginning, the stuntman runs horizontally off the roof.Therefore, the angle 0o = 0°.The horizontal distance is x = 4.5 m.The vertical distance is y = -1.9 m.Let the initial velocity be vo.Expression to find the initial speed of the stuntman (vo) is shownbelow:gxvo = ±V- (1)Here, gis acceleration due to gravity.Refer “Appendix E" in the textbook for the gravity value of theEarth.The value of gravity is g = 9.81 m/s².Substitute 9.81 m/s² for g, 4.5 m for x, and – 1.9 m for y in Equation(1).(9.81 m/s²) (4.5 m)*2(-1.9 m)Vo = ±= +7.2 m/sThe positive and negative sign of the result indicates the stuntmanmay runoff the building to the left or right.Therefore, the running speed of the stuntman is vo = ±7.2 m/s.

Solution: to find the relation of the x and y components of the v2 = v0^2 +2as or something else…

Answered: The Gap between the buildings is 4.5 m…

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter2: Vectors

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Good day to you

I was looking through a particular problem found in Richard Wolfson Essential University Physics Volume one, 4th Edition textbook, on chapter 3, problem 68. Your solution derives an inital speed equation, however, I don't know how it was derived. was it a manipulation of the x and y components of the v² = v₀^2 +2as or something else entirely?

Thanks in advance

< Chapter 3, Problem 68P
The Gap between the buildings is 4.5 m wide.
The stuntman runs from one building roof to another building
roof. The height between two building roofs is 1.9 m.
Calculation:
At the beginning, the stuntman runs horizontally off the roof.
Therefore, the angle 0o = 0°.
The horizontal distance is x = 4.5 m.
The vertical distance is y = -1.9 m.
Let the initial velocity be vo.
Expression to find the initial speed of the stuntman (vo) is shown
below:
gx
vo = ±V- (1)
Here, gis acceleration due to gravity.
Refer “Appendix E" in the textbook for the gravity value of the
Earth.
The value of gravity is g = 9.81 m/s².
Substitute 9.81 m/s² for g, 4.5 m for x, and – 1.9 m for y in Equation
(1).
(9.81 m/s²) (4.5 m)*
2(-1.9 m)
Vo = ±
= +7.2 m/s
The positive and negative sign of the result indicates the stuntman
may runoff the building to the left or right.
Therefore, the running speed of the stuntman is vo = ±7.2 m/s.

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