Homework 02 Solutions

Homework 02 Solutions - homework 02 – GADHIA, TEJAS –...

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Unformatted text preview: homework 02 – GADHIA, TEJAS – Due: Jan 31 2007, 4:00 am 1 Question 1 part 1 of 2 10 points The tallest volcano in the solar system is the 24 km tall Martian volcano, Olympus Mons. Assume: An astronaut drops a ball off the rim of the crater and that the free fall acceler- ation of the ball remains constant throughout the ball’s 24 km fall at a value of 3 . 2 m / s 2 . (We assume that the crater is as deep as the volcano is tall, which is not usually the case in nature.) a) Find the time for the ball to reach the crater floor. Correct answer: 122 . 474 s (tolerance ± 1 %). Explanation: Let : h = 24 km and a = 3 . 2 m / s 2 . Basic Concepts: Free fall s = s o + v t + 1 2 a t 2 v 2 = v 2 + 2 a h . Solution: The distance an object falls from rest under an acceleration a is h = 1 2 a t 2 , so t = radicalbigg 2 h a = radicalBigg 2 (24 km) (3 . 2 m / s 2 ) = radicalBig (153600 m 2 / s 2 ) = 122 . 474 s . Question 2 part 2 of 2 10 points b) Find the magnitude of the velocity with which the ball hits the crater floor. Correct answer: 391 . 918 m / s (tolerance ± 1 %). Explanation: Since the object falls from rest, v 2 = 2 a h so v = √ 2 a h = radicalBig 2 (3 . 2 m / s 2 ) (24 km) = radicalBig (153600 m 2 / s 2 ) = 391 . 918 m / s . Question 3 part 1 of 1 10 points A stone is thrown straight up- ward and at the top of its trajec- tory its velocity is momentarily zero. What is its acceleration at this point? 1. Zero 2. 9.8 m/s 2 up 3. 9.8 m/s 2 down correct 4. Unable to determine Explanation: Basic Concepts: The gravitational accel- eration near the surface of the earth is consid- ered constant, for all practical purposes. This acceleration of 9.8 m/s 2 is pointing downward. Solution: To illustrate how it works, let’s take, for example, an upward initial velocity of 9.8 m/s. One second later the velocity will be zero. One second after that it will be at- 9 . 8 m/s. In other words, in each second the velocity is decreased by 9.8 m/s. homework 02 – GADHIA, TEJAS – Due: Jan 31 2007, 4:00 am 2 Question 4 part 1 of 1 10 points The acceleration due to gravity on planet X is one fifth that on the surface of the earth. If it takes 4 . 2 s for an object to fall a certain distance from rest on earth, how long would it take to fall the same distance on planet X? Correct answer: 9 . 39149 s (tolerance ± 1 %). Explanation: Because the acceleration due to gravity is uniform near the surface of both planets, the distance an object falls in a time t is given by x fall = g planet t 2 2 where g planet is the force of gravity near the surface of the given planet. Since the distance x fall is the same for both planets, then x fall = g e t 2 e 2 = g x t 2 x 2 = 1 5 g e t 2 x 2 t 2 e = 1 5 t 2 x t x = √ 5 t e = √ 5 (4 . 2 s) = 9 . 39149 s ....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Homework 02 Solutions - homework 02 – GADHIA, TEJAS –...

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