Homework Solutions 6

# Homework Solutions 6 - homework 02 ASH BEN Due 11:00 pm...

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homework 02 – ASH, BEN – Due: Jan 30 2008, 11:00 pm 1 Question 1, chap 2, sect 6. 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. 9.8 m/s 2 up 2. Zero 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. Question 2, chap 2, sect 6. part 1 of 1 10 points The acceleration due to gravity on planet X is one Ffth that on the surface of the earth. If it takes 4 . 7 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: 10 . 5095 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 . 7 s) = 10 . 5095 s . Question 3, chap 2, sect 6. part 1 of 4 10 points A mountain climber stands at the top of a 50.9 m cli± hanging over a calm pool of water. The climber throws two stones vertically 1.7 s apart and observes that they cause a single splash when they hit the water. The Frst stone has an initial velocity of +2.3 m/s. a) What will the velocity of the Frst stone be at the instant both stones hit the water? Correct answer: - 31 . 6851 m / s (tolerance ± 1 %). Explanation: Basic Concept: v 2 f = v 2 i + 2 a Δ y Given: Δ y = - 50 . 9 m v i, 1 = 2 . 3 m / s Δ t 1 = Δ t 2 + 1 . 7 s a = - 9 . 81 m / s 2 Solution: v 2 f, 1 = (2 . 3 m / s) 2 + 2 ( - 9 . 81 m / s 2 ) ( - 50 . 9 m) = 1003 . 95 m 2 / s 2 v f, 1 = ± r 1003 . 95 m 2 / s 2 = ± 31 . 6851 m / s

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homework 02 – ASH, BEN – Due: Jan 30 2008, 11:00 pm 2 The velocity is - 31 . 6851 m / s since the motion is downward. Question 4, chap 2, sect 6. part 2 of 4 10 points b) How long after the release of the Frst stone will the two stones hit the water? Correct answer: 3 . 46434 s (tolerance ± 1 %). Explanation: Basic Concept: v f = v i + a Δ t Solution: Δ t 1 = v f, 1 - v i, 1 a = - 31 . 6851 m / s - (2 . 3 m / s) - 9 . 81 m / s 2 = 3 . 46434 s Question 5, chap 2, sect 6. part 3 of 4
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Homework Solutions 6 - homework 02 ASH BEN Due 11:00 pm...

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