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During a test a rocket travels upward at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height SB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s² due to gravity. Neglect the effect of air resistance.

During a test a rocket travels upward at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height SB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s² due to gravity. Neglect the effect of air resistance.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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This is from my engineering mechanics book
I'm not that familiar with limits yet in integral calculus, and I'm not that good with advanced calculus yet

and in the image is a solved problem

but I don't understand the steps on how it got to the answer

can you explain the  steps to me and I think there were shortcuts made in the solution can you give me the longer version or tell me any fundamental formulas I should already know when solving this  kind of problem, so I can understand how the book derived to the answer.

VA= 75 m/s SA UB=0 = 40 m B Fig. 12-4 SB. VB = 0. For the entire motion, the acceleration is a = -9.81 m/s² (negative since it acts in the opposite sense to positive velocity or positive displacement). Since a, is constant the rocket's position may be related to its velocity at the two points A and B on the path by using Eq. 12-6, namely, (+1) v=v²A + 2a(SB - SA) 0 = (75 m/s)² + 2(-9.81 m/s²) (SB- 40 m) SB 327 m Ans. Velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply Eq. 12-6 between points B and C, Fig. 12-4. (+1) v = v + 2a(sc - SB) = 0 + 2(-9.81 m/s²) (0 - 327 m) Vc = -80.1 m/s = 80.1 m/s ↓ The negative root was chosen since the rocket is moving downward. Similarly, Eq. 12-6 may also be applied between points A and C, i.e., (+1) v=v² + 2a(sc-SA) = = (75 m/s)2 + 2(-9.81 m/s²) (0 - 40 m) vc = -80.1 m/s = 80.1 m/s ↓ Ans. Ans. NOTE: It should be realized that the rocket is subjected to a deceleration from A to B of 9.81 m/s², and then from B to C it is accelerated at this rate. Furthermore, even though the rocket momentarily comes to rest at B (VB = 0) the acceleration at B is still 9.81 m/s² downward!
Transcribed Image Text:VA= 75 m/s SA UB=0 = 40 m B Fig. 12-4 SB. VB = 0. For the entire motion, the acceleration is a = -9.81 m/s² (negative since it acts in the opposite sense to positive velocity or positive displacement). Since a, is constant the rocket's position may be related to its velocity at the two points A and B on the path by using Eq. 12-6, namely, (+1) v=v²A + 2a(SB - SA) 0 = (75 m/s)² + 2(-9.81 m/s²) (SB- 40 m) SB 327 m Ans. Velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply Eq. 12-6 between points B and C, Fig. 12-4. (+1) v = v + 2a(sc - SB) = 0 + 2(-9.81 m/s²) (0 - 327 m) Vc = -80.1 m/s = 80.1 m/s ↓ The negative root was chosen since the rocket is moving downward. Similarly, Eq. 12-6 may also be applied between points A and C, i.e., (+1) v=v² + 2a(sc-SA) = = (75 m/s)2 + 2(-9.81 m/s²) (0 - 40 m) vc = -80.1 m/s = 80.1 m/s ↓ Ans. Ans. NOTE: It should be realized that the rocket is subjected to a deceleration from A to B of 9.81 m/s², and then from B to C it is accelerated at this rate. Furthermore, even though the rocket momentarily comes to rest at B (VB = 0) the acceleration at B is still 9.81 m/s² downward!
VA75 m/s 1 A UB=0 SA = 40 m B SB During a test a rocket travels upward at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height SB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s² due to gravity. Neglect the effect of air resistance. SOLUTION Coordinate System. The origin O for the position coordinate s is taken at ground level with positive upward, Fig. 12-4. Maximum Height. Since the rocket is traveling upward, VA +75m/s when / = 0. At the maximum height s = Sg the velocity VB = 0. For the entire motion, the acceleration is a = -9.81 m/s² (negative since it acts in the opposite sense to positive velocity or positive displacement). Since a, is constant the rocket's position may be related to its velocity at the two points A and B on the path by using Eq. 12-6, namely, (+1) v² = v²A + 2ac($B - SA) 0 SB = 327 m Ans. Velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply Eq. 12-6 between points B and C, Fig. 12-4. (+1) v = v + 2a(sc - SB) = 0 + 2(-9.81 m/s²) (0 - 327 m) vc-80.1 m/s = 80.1 m/s (75 m/s)² + 2(-9.81 m/s²) (SB- 40 m) Ans.
Transcribed Image Text:VA75 m/s 1 A UB=0 SA = 40 m B SB During a test a rocket travels upward at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height SB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s² due to gravity. Neglect the effect of air resistance. SOLUTION Coordinate System. The origin O for the position coordinate s is taken at ground level with positive upward, Fig. 12-4. Maximum Height. Since the rocket is traveling upward, VA +75m/s when / = 0. At the maximum height s = Sg the velocity VB = 0. For the entire motion, the acceleration is a = -9.81 m/s² (negative since it acts in the opposite sense to positive velocity or positive displacement). Since a, is constant the rocket's position may be related to its velocity at the two points A and B on the path by using Eq. 12-6, namely, (+1) v² = v²A + 2ac($B - SA) 0 SB = 327 m Ans. Velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply Eq. 12-6 between points B and C, Fig. 12-4. (+1) v = v + 2a(sc - SB) = 0 + 2(-9.81 m/s²) (0 - 327 m) vc-80.1 m/s = 80.1 m/s (75 m/s)² + 2(-9.81 m/s²) (SB- 40 m) Ans.
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