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hw17 - Muraj Hamza Homework 17 Due Mar 3 2006 4:00 am Inst...

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Muraj, Hamza – Homework 17 – Due: Mar 3 2006, 4:00 am – Inst: Florin 1 This print-out should have 12 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 2) 10 points A rock of mass m is thrown horizontally off a building from a height h . The speed of the rock as it leaves the thrower’s hand at the edge of the building is v 0 as shown. v 0 h Δ x How much time does it take the rock to travel from the edge of the building to the ground? 1. t = 2 h g 2. t = h v 0 g 3. t = s 2 h g correct 4. t = h v 0 5. t = p h v 0 Explanation: The motion of the rock in the y -direction is a free falling motion from rest: h = 1 2 g t 2 t = s 2 h g . 002 (part 2 of 2) 10 points What is the kinetic energy of the rock just before it hits the ground? 1. K f = m g h - 1 2 m v 2 0 2. K f = 1 2 m v 2 0 - m g h 3. K f = 1 2 m v 2 0 + m g h correct 4. K f = 1 2 m v 2 0 5. K f = m g h Explanation: By the work-energy theorem, K f - K i = W , where in the present problem, W = ( m g ) h and K i = 1 2 m v 2 0 . Thus, K f = K i + W = 1 2 m v 2 0 + m g h . 003 (part 1 of 1) 10 points A projectile of mass 0 . 271 kg is shot from a cannon, at height 6 . 6 m, as shown in the figure, with an initial velocity v i having a horizontal component of 5 . 9 m / s. The projectile rises to a maximum height of Δ y above the end of the cannon’s barrel and strikes the ground a horizontal distance Δ x past the end of the cannon’s barrel. The acceleration of gravity is 9 . 8 m / s 2 . Δ x v i 49 Δ y 6 . 6 m Find the magnitude of the final velocity vector when the projectile hits the ground. Correct answer: 14 . 4995 m / s. Explanation: Let : v x i = 5 . 9 m / s , x i = 0 m , y i = 6 . 6 m , and θ = 49 . For the horizontal component of vector v i ,
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Muraj, Hamza – Homework 17 – Due: Mar 3 2006, 4:00 am – Inst: Florin 2 we have v x i = v i cos θ ; consequently v i = v x i cos θ = (5 . 9 m / s) cos 49 = 8 . 99309 m / s . The work done by gravity depends only on the vertical distance y i , since the work involving Δ y adds and subtracts; i.e. , cancels. The work done by gravity is W = m g y i .
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