Homework 4

# Homework 4 - homework 04 BAUTISTA ALDO Due 4:00 am Version...

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homework 04 – BAUTISTA, ALDO – Due: Jan 30 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. A ball is thrown upward. Its initial verti- cal speed is 8 . 6 m / s and the acceleration of gravity is 9 . 8 m / s 2 , as shown in the Fgure below. Neglect: Air resistance. 8 . 6m / s 9 8m 2 h max What is its maximum height, h max ? Correct answer: 3 . 77347 m (tolerance ± 1 %). Explanation: Let : v 0 = 8 . 6 m / s and g = 9 . 8 m / s 2 . Basic Concept: ±or constant accelera- tion, we have v 2 = v 2 0 + 2 a ( x - x 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 = (8 . 6 m / s) 2 2 (9 . 8 m / s 2 ) = 3 . 77347 m . Question 2 Part 2 of 2. 10 points. ±ind the speed v A of the ball as the ball passes a point A , which is at one quarter of the maximum height h max 4 . Correct answer: 7 . 44782 m / s (tolerance ± 1 %). Explanation: At the point A , the velocity is again found from the same equation, but with the Fnal velocity v A v 2 A = v 2 0 + 2 ( - g ) h 4 h = v 2 0 2 g from Part 1, so h 4 = v 2 0 8 g and v 2 A = v 2 0 - 2 g ± v 2 0 8 g ² = v 2 0 - v 2 0 4 = 3 4 v 2 0 . Thus v A = 3 2 v 0 = 3 2 (8 . 6 m / s) = 7 . 44782 m / s . Question 3 Part 1 of 2. 10 points. An airplane ²ying parallel to the ground undergoes two consecutive displacements. The Frst is 73 km at 27 . 9 west of north, and the second is 160 km at 69 . 3 east of north. a) What is the magnitude of the plane’s total displacement? Correct answer: 167 . 336 km (tolerance ± 1 %).

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homework 04 – BAUTISTA, ALDO – Due: Jan 30 2006, 4:00 am 2 Explanation: 73 km 117 . 9 160 20 . 7 d Note: Figure is not drawn to scale. Basic Concepts: Δ x = d (sin θ ) Δ y = d (cos θ ) Δ x total = Δ x 1 + Δ x 2 Δ y total = Δ y 1 + Δ y 2 d total = q x total ) 2 + (Δ y total ) 2 Given: d 1 = 73 km θ 1 = 90 . 0 + 27
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Homework 4 - homework 04 BAUTISTA ALDO Due 4:00 am Version...

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