HW5.sol - homework 05 – RAHMAN TARIQUE – Due 3:00 am 1...

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Unformatted text preview: homework 05 – RAHMAN, TARIQUE – Due: Feb 11 2008, 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 R ? 1. 2. 3. 4. The ball is in free-fall and there is no acceleration at any point on its path. 5. correct 6. 7. 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 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 How do the speeds of the ball at the three points compare? 1. bardbl vectorv P bardbl < bardbl vectorv Q bardbl < bardbl vectorv R bardbl 2. bardbl vectorv P bardbl = bardbl vectorv R bardbl = bardbl vectorv Q bardbl 3. bardbl vectorv Q bardbl < bardbl vectorv P bardbl = bardbl vectorv R bardbl correct 4. bardbl vectorv P bardbl = bardbl vectorv R bardbl < bardbl vectorv Q bardbl 5. bardbl vectorv R bardbl < bardbl vectorv Q bardbl < bardbl vectorv P bardbl 6. bardbl vectorv Q bardbl < bardbl vectorv R bardbl < bardbl vectorv P bardbl Explanation: The speed of the ball in the x-direction is constant. Because of gravitational accelera- tion, the speed in the y-direction is zero at point Q . Since points P and R are located at the same point above ground, by symmetry we see that they have the same vertical speed component (though they do not have the same homework 05 – RAHMAN, TARIQUE – Due: Feb 11 2008, 3:00 am 2 velocity). The answer is then “ v Q < v P = v R ”. Question 3, chap 4, sect 4. part 1 of 3 10 points A projectile is fired with an initial speed v at t = 0. The angle between the initial velocity v and the horizontal plane is α . x y v α y max R A B The time t max it takes for the projectile to reach its maximum height is 1. t max = v cos α g 2. t max = v cos α 2 g 3. t max = v g 4. t max = g v 5. t max = v sin α g correct 6. t max = v g 7. t max = v 8. t max = v sin α 2 g Explanation: Basic Concepts: For two dimensional projectile motion in a gravitational field the acceleration is due to gravity only and acts exclusively on the y component of velocity....
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HW5.sol - homework 05 – RAHMAN TARIQUE – Due 3:00 am 1...

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