Kaplunovsky - hw 03

# Kaplunovsky hw 03 - homework 03 – – Due Feb 7 2007 4:00 am 1 Question 1 chap 2 sect 6 part 1 of 1 5 points An object is shot vertically upward

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Unformatted text preview: homework 03 – – Due: Feb 7 2007, 4:00 am 1 Question 1, chap 2, sect 6. part 1 of 1 5 points An object is shot vertically upward into the air with a positive initial velocity. Which of the following correctly describes the velocity and acceleration of the object at its maximum elevation? Velocity Acceleration 1. positive positive 2. negative negative 3. positive negative 4. zero negative correct 5. zero zero Explanation: At the maximum elevation, the vertical ve- locity is zero. The acceleration is due to gravity, which acts down. Question 2, chap 2, sect 6. part 1 of 1 5 points An object is thrown upward from the origin along the positive y direction and caught. Which graph correctly describes the y coor- dinate of the vertical motion while the object is in the air? 1. t y 2. t y 3. t y 4. t y 5. None of these graphs are correct. 6. t y 7. t y 8. t y correct 9. t y Explanation: The vertical motion should be represented by the parabola y = v t − 1 2 g t 2 , as shown t y homework 03 – – Due: Feb 7 2007, 4:00 am 2 Question 3, chap 2, sect 6. part 1 of 1 5 points An object is released from rest on a planet that has no atmosphere. The object falls freely for 5 . 89 m in the first second. What is the magnitude of the acceleration due to gravity on the planet? Correct answer: 11 . 78 m / s 2 (tolerance ± 1 %). Explanation: Let : s = 5 . 89 m . s = 1 2 a t 2 a = 2 s t 2 = 2 (5 . 89 m) (1 s) 2 = 11 . 78 m / s 2 . Question 4, chap 2, sect 6. part 1 of 1 7 points An apple falls from a tree and hits the ground 11 . 2 m below. The acceleration of gravity is 9 . 8 m / s 2 . With what speed will it hit the ground? Correct answer: 14 . 8162 m / s (tolerance ± 1 %). Explanation: When an object undergoes free fall from rest, its final speed is given by v = radicalbig 2 gh Question 5, chap 2, sect 5. part 1 of 1 7 points The barrel of a rifle has a length of 0 . 823 m. A bullet leaves the muzzle of a rifle with a speed of 553 m / s. Note: A bullet in a rifle barrel does not have constant acceleration, however, constant acceleration is assumed for this problem. What is the acceleration of the bullet while in the barrel? Correct answer: 185789 m / s 2 (tolerance ± 1 %). Explanation: The initial velocity v i = 0. Under the assumption that the acceleration of the bullet was constant while in the barrel, v 2 f = v 2 i + 2 a d = 2 a ℓ . Thus a = v 2 f 2 ℓ = (553 m / s) 2 2 (0 . 823 m) = 185789 m / s 2 . Question 6, chap 2, sect 6. part 1 of 1 8 points The acceleration due to gravity on the moon is about one-sixth its value on earth. If a baseball reaches a height of 63 m when thrown upward by someone on the earth, what height would it reach when thrown in the same way on the surface of the moon?...
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## This note was uploaded on 01/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Kaplunovsky hw 03 - homework 03 – – Due Feb 7 2007 4:00 am 1 Question 1 chap 2 sect 6 part 1 of 1 5 points An object is shot vertically upward

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