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Unformatted text preview: homework 03 – ARMSTRONG, DOMINIC – Due: Sep 25 2007, 3:00 am 1 Question 1, chap 4, sect 4. part 1 of 1 10 points A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Q R P Which of the following diagrams best indi cates the direction of the acceleration, if any, on the ball at point P ? 1. 2. 3. 4. 5. 6. correct 7. The ball is in freefall and there is no acceleration at any point on its path. 8. 9. Explanation: Since air friction is negligible, the only ac celeration on the ball after being thrown is that due to gravity, which acts straight down. Question 2, chap 4, sect 4. part 1 of 1 10 points A cannon fires a 0 . 784 kg shell with initial velocity v i = 10 m / s in the direction θ = 56 ◦ above the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . Δ x Δ h 1 m / s 5 6 ◦ Δ y y The shell’s trajectory curves downward be cause of gravity, so at the time t = 0 . 62 s the shell is below the straight line by some vertical distance. Your task is to calculate this distance Δ h in the absence of air resistance. Correct answer: 1 . 88356 m (tolerance ± 1 %). Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: ˆ x = tv i cos θ , ˆ y = tv i sin θ . The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate at constant downward acceleration a y = − g , homework 03 – ARMSTRONG, DOMINIC – Due: Sep 25 2007, 3:00 am 2 hence x = tv i cos θ, y = tv i sin θ − g t 2 2 . Thus, x = ˆ x but y = ˆ y − 1 2 g t 2 , or in other words, the shell deviates from the straightline path by the vertical distance ∆ h = ˆ y − y = g t 2 2 . Note: This result is completely indepen dent on the initial velocity v i or angle θ of the shell. It is a simple function of the flight time t and nothing else (besides the constant g = 9 . 8 m / s 2 ). ∆ h = g t 2 2 = (9 . 8 m / s 2 ) (0 . 62 s) 2 2 = 1 . 88356 m . Question 3, chap 4, sect 4. part 1 of 1 10 points For your information: g = 9 . 8 m / s 2 . A car is parked near a cliff overlooking the ocean on an incline that makes an angle θ = 17 . 1 ◦ with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has velocity 5 m / s when it reaches the cliff’s edge, 44 m above the ocean. When the car hits the water, how far is it from the cliff’s base? θ Correct answer: 13 . 6216 m (tolerance ± 1 %). Explanation: First, let’s find out the flight time of the car. Consider the vertical motion of the car — it has constant acceleration g . Initially, when the car goes off the cliffs edge, its vertical velocity is v y = v × sin θ = 1 . 4702 m / s ....
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This homework help was uploaded on 04/07/2008 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Swinney
 mechanics, Friction, Work

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