A projectile is fired at an upward angle of 45.0° from the

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Problem 38P Chapter 6

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 38P

A projectile is fired at an upward angle of 45.0° from the top of a 265-m cliff with a speed of 185 m/s. What will be its speed when it strikes the ground below? (Use conservation of energy.)

Step-by-Step Solution:

Solution Step 1 of 5 Asteroid belt Asteroid belt is the circular disc located between the orbits of the mars and the jupiter planets, which contains numerous irregular shaped bodies called asteroids or minor planets. As shown in figure below The quantities like orbital period, speed and acceleration of the planet or asteroids revolving around the sun, does not depend on the mass of the planet or asteroid. It depends on the mass of the central body around which the planet is revolving and the radius of the orbit. Orbital time period The time period in the given orbit is the time taken by the space shuttle to complete one full revolution around earth. The relation between the orbital time period T and the distance of space shuttle r from the center of the earth, 2 T = 4 r GM Where T is the orbital time period, r is the radius of the orbit, G is gravitational constant and M is the mass of the sun. From the above equation, the radius of the orbit, 2 1/3 r = {GM42 } Step 2 of 5 Given data One of the asteroid in the asteroid belt has the time period of 5 earth years That is, T= 5 years Converting years to seconds, Using 1 year=365 days 1 day= 24 hours 1 hour= 3600 seconds 365 days 24 hour 3600 sec T= 5 years( 1 year)( 1 day)( 1 hour) T = 15768×10 sec 4 Mass of the sun, M=1.99×10 kg 30 G=6.67 ×10 11 Nm /kg 2 To find, Radius of the asteroid orbit, r= Step 3 of 5 Using given data in above equation of radius of orbit, 11 2 2 30 4 2 r = { (6.67 ×10 Nm /kg )(1.99×10 kg)(15768×10 se})1/3 4(3.14) r = 4.38×10 m 11 Therefore, the orbital radius of the given asteroid around the sun is 4.38×10 m. 11

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Chapter 6, Problem 38P is Solved
Step 5 of 5

Textbook: Physics: Principles with Applications
Edition: 6th
Author: Douglas C. Giancoli
ISBN: 9780130606204

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