Explanation From conservation of energy, energy before launch is equal to energy after launched Write the expression for energy of the projectile before and after launch using conservation of energy. K i + U i + Δ E mech = K f + U f (I) Here, K i is the kinetic energy of the projectile before the launch, U i is the potential energy of the projectile before the launch, Δ E mech is the change in mechanical energy of the projectile, K f is the kinetic energy of the projectile before the launch, and U f is the potential energy of the projectile after the launch. The height to which the projectile is projected is compared with respect to the radius of earth. Write the expression for gravitational energy of the projectile before the launch. U i = G M E M P R E (II) Here, U i is the gravitational energy of the projectile before the launch, G is the universal gravitational constant, M E is the mass of earth, M P is the mass of the projectile , and R E is the radius of the earth. Write the expression for gravitational energy of the projectile at a height h . U f = G M E M P R E + h (III) Here, U f is the gravitational energy of the projectile at a height h , and h is the height to which the projectile is projected. Write the expression for initial kinetic energy of the projectile. K i = 1 2 M p v i 2 (IV) Here, v i is the initial velocity of the projectile. Use equation (II), (III) and (IV) in equation (I). 1 2 M p v i 2 + G M E M P R E + Δ E mech = K f + G M E M P R E + h (V) The change in mechanical energy and the final kinetic energy of the projectile is zero, thus the above equation is reduced to 1 2 M p v i 2 + G M E M P R E = G M E M P R E + h (VI) Rearrange equation (VI), to find h . 1 2 M p v i 2 + G M E M P R E = G M E M P R E + h ( 1 2 M p v i 2 + G M E M P R E + h ) = G M E M P R E 1 2 v i 2 − G M E R E = − G M E R E + h (VII) Conclusion: Substitute 6

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN: 9781133104261

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Chapter 6, Problem 53P

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The height to which the projectile rises by ignoring air resistance.

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Chapter 6, Problem 52P

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Principles of Physics: A Calculus-Based Text

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