PHY302K-Sai-HW2-solutions

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

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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 – fnd all choices beFore answering. 001 10.0 points A ball is thrown upward. Its initial verti- cal speed is 11 . 1m / s , acceleration oF gravity is 9 . 8m / s 2 , and maximum height h max are shown in the fgure below. Neglect: Air resistance. 11 . 1m / s 9 8m 2 h max What is its maximum height, h max ? Correct answer: 6 . 28622 m. Explanation: Let : v 0 = 11 . 1m / s and g =9 . 8m / s 2 . Basic Concept: ±or 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 . 1m / s) 2 2 (9 . 8m / 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 . 8m / s 2 . t h y 2 . 9s 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 gt 2 . Let : t =2 . 9s , v o = 19 m / s , and g =9 . 8m / s 2 . Solution: h 1 = - vt - 1 2 gt 2 = - (19 m / s) (2 . 9 s) - 1 2 (9 . 8m / 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 . 9s 19 m / s b) up with an initial speed of 19 m / s? Correct answer: 13 . 891 m. Explanation: h 2 = vt - 1 2 gt 2 = (19 m / s) (2 . 9 s) - 1 2 (9 . 8m / s 2 ) (2 . 9 s) 2 = 13 . 891 m . 004 10.0 points You and your friend throw balloons Flled 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 . 8m / s 2 . You can neglect air resistance. How high is the apartment house? Correct answer: 150 . 932 m. Explanation: ±or balloon 1: y 1 = 1 2 gt 2 1 ±or balloon 2: y 2 = v 0 t 2 + gt 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 gt 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 ±or Δ t = 2 s t 1 = 1 + v 0 v 0 - 2 g ±or v 0 = 30 m/s, Δ t = 2 s t 1 =5 . 55 s ±rom the above, once t 1 is known, the height of the house is given simply by y = gt 2 1 2 ±or t 1 =5 . 55 s, y = 150 . 932 m 005 10.0 points
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PHY302K-Sai-HW2-solutions - Oslund (hno56) Homework 2 sai...

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