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Learning Goal: To apply the law of conservation of energy to an object launched upward in Earth's gravitational field. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K = (1/2)mv² and its gravitational potential energy U = mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation K₁ + U₁ = Kf +Uf where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. ▼ Part E At what height h above the ground does the projectile have a speed of 0.5v? Express your answer in terms of v and g. You may or may not use all of these quantities. h = Submit Part F ΠΗΓΙ ΑΣΦ V U = Request Answer What is the speed u of the object at the height of (1/2)hmax? Express your answer in terms of u and g. You may or may not use all of these quantities. ► View Available Hint(s) Submit ? VE ΑΣΦ ? First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance.

Learning Goal: To apply the law of conservation of energy to an object launched upward in Earth's gravitational field. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K = (1/2)mv² and its gravitational potential energy U = mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation K₁ + U₁ = Kf +Uf where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. ▼ Part E At what height h above the ground does the projectile have a speed of 0.5v? Express your answer in terms of v and g. You may or may not use all of these quantities. h = Submit Part F ΠΗΓΙ ΑΣΦ V U = Request Answer What is the speed u of the object at the height of (1/2)hmax? Express your answer in terms of u and g. You may or may not use all of these quantities. ► View Available Hint(s) Submit ? VE ΑΣΦ ? First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance.

An Introduction to Physical Science
An Introduction to Physical Science
14th Edition
ISBN:9781305079137
Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Chapter2: Motion
Section2.5: Projectile Motion
Problem 1PQ: Neglecting air resistance, why would a ball projected horizontally and another ball dropped at the...
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Learning Goal: To apply the law of conservation of energy to an object launched upward in Earth's gravitational field. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K = (1/2)mv² and its gravitational potential energy U = mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation K₁ + U₁ = Kf +Uf where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. ▼ Part E At what height h above the ground does the projectile have a speed of 0.5v? Express your answer in terms of v and g. You may or may not use all of these quantities. h = Submit Part F ΠΗΓΙ ΑΣΦ V U = Request Answer What is the speed u of the object at the height of (1/2)hmax? Express your answer in terms of u and g. You may or may not use all of these quantities. ► View Available Hint(s) Submit ? VE ΑΣΦ ?
Transcribed Image Text:Learning Goal: To apply the law of conservation of energy to an object launched upward in Earth's gravitational field. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K = (1/2)mv² and its gravitational potential energy U = mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation K₁ + U₁ = Kf +Uf where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. ▼ Part E At what height h above the ground does the projectile have a speed of 0.5v? Express your answer in terms of v and g. You may or may not use all of these quantities. h = Submit Part F ΠΗΓΙ ΑΣΦ V U = Request Answer What is the speed u of the object at the height of (1/2)hmax? Express your answer in terms of u and g. You may or may not use all of these quantities. ► View Available Hint(s) Submit ? VE ΑΣΦ ?
First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance.
Transcribed Image Text:First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance.
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