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Homework 2 Solutions

# Homework 2 Solutions - homework 02 ALIBHAI ZAHID Due 4:00...

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homework 02 – ALIBHAI, ZAHID – 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 32 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 32 km fall at a value of 2 . 7 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: 153 . 96 s (tolerance ± 1 %). Explanation: Let : h = 32 km and a = 2 . 7 m / s 2 . Basic Concepts: Free fall s = s o + v 0 t + 1 2 a t 2 v 2 = v 2 0 + 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 (32 km) (2 . 7 m / s 2 ) = radicalBig (172800 m 2 / s 2 ) = 153 . 96 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: 415 . 692 m / s (tolerance ± 1 %). Explanation: Since the object falls from rest, v 2 = 2 a h so v = 2 a h = radicalBig 2 (2 . 7 m / s 2 ) (32 km) = radicalBig (172800 m 2 / s 2 ) = 415 . 692 m / s . Question 3 part 1 of 2 10 points A ball is thrown upward. After reaching a maximum height, it continues falling back towards Earth. On the way down, the ball is caught at the same height at which it was thrown upward. Neglect: Air resistance. The acceleration of gravity is 9 . 8 m / s 2 . Its initial vertical speed v 0 , acceleration of gravity g , and maximum height h max are shown in the figure below. v 0 9 . 8 m / s 2 h max If the time ( up and down ) the ball remains in the air is t , calculate its speed v f when it caught. 1. v f = 2 g t 2. v f = 2 g t 3. v f = 1 2 g t 4. v f = g t

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homework 02 – ALIBHAI, ZAHID – Due: Jan 31 2007, 4:00 am 2 5. v f = 1 4 g t 6. v f = 1 2 g t correct 7. v f = 4 g t Explanation: Basic Concept: For constant accelera- tion, we have v 0 = g t (1) y = y 0 + v 0 t + 1 2 a t 2 . (2) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing h max = y y 0 , a , and t . Choose the positive direction to be up, then a = g and t up = v 0 g . From symmetry, t down = t up , we have t total = 2 v 0 g , or v f = v 0 = 1 2 g t total . Question 4 part 2 of 2 10 points If the time the ball remains in the air is t , calculate the maximum height h max the ball attained while in the air. 1. h max = 4 g t 2 2. h max = 1 4 g t 2 3. h max = 8 g t 2 4. h max = 1 8 g t 2 correct 5. h max = g t 2 6. h max = 1 2 g t 2 7. h max = 2 g t 2 Explanation: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 2 con- taining h max = y y 0 , a , and t . Choose the positive direction to be up, then a = g and h max = v 0 t 1 2 g t 2 = g t 2 1 2 g t 2 = 1 2 g t 2 , where v 0 = g t . Therefore t up = radicalbigg 2 y max g . From symmetry, t down = t up , we have t total = 2 radicalbigg 2 y max g .
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