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PHY302K-Sai-HW2-solutions

PHY302K-Sai-HW2-solutions - Oslund(hno56 Homework 2...

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Oslund (hno56) – Homework 2 – sai – (58015) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A ball is thrown upward. Its initial verti- cal speed is 11 . 1 m / s , acceleration of gravity is 9 . 8 m / s 2 , and maximum height h max are shown in the figure below. Neglect: Air resistance. 11 . 1 m / s 9 . 8 m / s 2 h max What is its maximum height, h max ? Correct answer: 6 . 28622 m. Explanation: Let : v 0 = 11 . 1 m / s and g = 9 . 8 m / s 2 . Basic Concept: For constant accelera- tion, we have v 2 = v 2 0 + 2 a ( y - y 0 ) . (1) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing v , a , y 0 , and y (no time) is the easiest one to use. Choose the positive direction to be up; then a = - g and 0 = v 2 0 + 2 ( - g ) ( h max - 0) or h max = v 2 0 2 g = (11 . 1 m / s) 2 2 (9 . 8 m / s 2 ) = 6 . 28622 m . 002 (part 1 of 2) 10.0 points A ball has an initial speed of 19 m / s. The acceleration of gravity is 9 . 8 m / s 2 . t h y 2 . 9 s 19 m / s What will be its position after 2 . 9 s if it is thrown a) down with an initial speed of 19 m / s? Correct answer: - 96 . 309 m. Explanation: Basic Concepts: This problem involves initial velocities directed both downward and upward, so all downward motion must be neg- ative. Assuming the initial position is h o = 0, the position at any time t is given by h f = v o t - 1 2 g t 2 . Let : t = 2 . 9 s , v o = 19 m / s , and g = 9 . 8 m / s 2 . Solution: h 1 = - v t - 1 2 g t 2 = - (19 m / s) (2 . 9 s) - 1 2 (9 . 8 m / s 2 ) (2 . 9 s) 2 = - 96 . 309 m .
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Oslund (hno56) – Homework 2 – sai – (58015) 2 003 (part 2 of 2) 10.0 points t h y 2 . 9 s 19 m / s b) up with an initial speed of 19 m / s? Correct answer: 13 . 891 m. Explanation: h 2 = v t - 1 2 g t 2 = (19 m / s) (2 . 9 s) - 1 2 (9 . 8 m / s 2 ) (2 . 9 s) 2 = 13 . 891 m . 004 10.0 points You and your friend throw balloons filled with water from the roof of a several story apart- ment house. You simply drop a balloon from rest. A second balloon is thrown downward by your friend 3 . 7 s later with an initial speed of 72 . 52 m / s. They hit the ground simultane- ously. The acceleration of gravity is 9 . 8 m / s 2 . You can neglect air resistance. How high is the apartment house? Correct answer: 150 . 932 m. Explanation: For balloon 1: y 1 = 1 2 g t 2 1 For balloon 2: y 2 = v 0 t 2 + g t 2 2 2 Balloon 2 is thrown Δ t seconds after balloon 1 and they hit the ground at the same time, so: t 1 = t 2 + Δ t Since the height of the house is the same for both of the balloons, y 1 = y 2 g t 2 1 2 = v 0 ( t 1 - Δ t ) + g 2 ( t 1 - Δ t ) 2 ( v 0 - g Δ t ) t 1 = v 0 Δ t - g 2 Δ t 2 t 1 = v 0 Δ t - g 2 Δ t 2 v 0 - g Δ t = Δ t 2 + v 0 v 0 - g Δ t Δ t 2 For Δ t = 2 s t 1 = 1 + v 0 v 0 - 2 g For v 0 = 30 m/s, Δ t = 2 s t 1 = 5 . 55 s From the above, once t 1 is known, the height of the house is given simply by y = g t 2 1 2 For t 1 = 5 . 55 s, y = 150 . 932 m 005 10.0 points Michael stands motionless holding a base- ball in his hand. After a while he tosses it upwards, and it travels up for a while be- fore turning about and heading towards the ground. Define upwards to be positive.
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