Question

17. If an object is launched from the Earth at two thirds of its escape velocity, what is the maximum height above the surface which the projectile reaches? A. 2.18 x 106 m B. 1.91 x 107 m C. 1.27 x 106 m D. 1.15 × 107 m E. 5.10 x 106 m 18. Consider a parachutist with their chute open descending from a plane. His terminal velocity is vy. Suddenly, the chute develops a hole reduces the coefficient of the drag force (C) acting on the parachutist to half its original value. The terminal velocity of the parachutist will now be A 2v, B. c. (42 1) D. V2v, E. 2

Answer 1

17. The escape velocity of Earth is given as follows: The velocity of projectile is given as follows: 2 2GM 31 R Use conservation of energy to calculate the value of projectile's maximum height as follows: The kinetic energy of the projectile at the surface of the Earth is given as follows: 4GMm 9R The potential energy of the projectile at surface of Earth is given as follows: GmM P= (R+h) The potential energy of the projectile at maximum height h is given as follows: P-GMm R+h Equate the Initial total energy and final total energy of projectile and rearrange for h as follows:

K+P. =P 4GMm GMM GMM 9RR (R+h) 4 11 9R R (R+h) -1 --R 74.1 (OR R Calculate the maximum height achieved by the projectile as follows: 4 11-R 9RR 9RR 5 4(6.371x10 m = 5.10x10m Therefore, projectile achieves maximum height of 5.1x109 m. 18. The formula for terminal velocity in terms of drag coefficient C is given as follows: V = 2mg mg V PAC If drag coefficient changes to half of its value, the terminal velocity is given as follows: =V2v Therefore, correct option is option-D.