# hw3sol - homework 03 – ARMSTRONG, DOMINIC – Due: Sep 25...

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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 free-fall 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 straight-line 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.

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hw3sol - homework 03 – ARMSTRONG, DOMINIC – Due: Sep 25...

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