This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Gozick, Brandon – Homework 4 – Due: Sep 19 2006, 7:00 pm – Inst: D Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A man can throw a ball a maximum horizontal distance of 90 . 3 m. The acceleration of gravity is 9 . 8 m / s 2 . How far can he throw the same ball verti- cally upward with the same initial speed? Correct answer: 45 . 15 m. Explanation: The range of a particle is given by the ex- pression R = v 2 sin2 θ g . The maximum horizontal distance is obtained when the ball is thrown at an angle θ = 45 ◦ and sin2 θ = 1 . Solving for v , v = p g R . When the ball is thrown upward with this speed, the maximum height is obtained from the equation v 2 f = v 2- 2 g h. Let v f = 0, and solve for h h = v 2 2 g . keywords: 002 (part 1 of 1) 10 points A ball is thrown from the top of a building upward at an angle of 35 ◦ to the horizontal and with an initial speed of 7 m / s, as in the figure. The ball is thrown at a height of 38 m above the ground and hits the ground 18 . 4893 m from the base of the building. The acceleration of gravity is 9 . 8 m / s 2 . 3 5 ◦ 38m ground level hand height 7 m / s 18 . 4893 m What is the speed of the ball just before it strikes the ground? Correct answer: 28 . 1745 m / s. Explanation: Let : θ = 35 ◦ , h = 38 m , and v = 7 m / s . Separate the velocity into its x component and y component....
View Full Document