Oldhw9_s - oldhomewk 09 – YOO HEE – Due Feb 5 2008 4:00 am 1 Question 1 chap 4 sect 6 part 1 of 3 10 points A skater(water spider maintains an

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Unformatted text preview: oldhomewk 09 – YOO, HEE – Due: Feb 5 2008, 4:00 am 1 Question 1, chap 4, sect 6. part 1 of 3 10 points A skater (water spider) maintains an av- erage position on the surface of a stream by darting upstream (against the current), then drifting downstream (with the current) to its original position. The current in the stream is 0 . 719 m / s relative to the shore, and the skater darts upstream 0 . 262 m (relative to a spot on shore) in 1 . 11 s during the first part of its motion. Take upstream as the positive direction. Determine the velocity of the skater relative to the water during its dash upstream. Correct answer: 0 . 955036 m / s (tolerance ± 1 %). Explanation: v sg = velocity of the skater relative to the ground, v sw = velocity of the skater relative to the water, and v wg = velocity of the water relative to the ground. vector v sg = vector v sw + vector v wg . (1) We have v wg = - 0 . 719 m / s, and v sg = l t = . 262 m 1 . 11 s = 0 . 236036 m / s . Equation (1) gives v sw = v sg- v wg = 0 . 236036 m / s- (- . 719 m / s) = 0 . 955036 m / s . Question 2, chap 4, sect 6. part 2 of 3 10 points How far upstream relative to the water does the skater move during one cycle of this mo- tion? Correct answer: 1 . 06009 m (tolerance ± 1 %). Explanation: d sw = v sw t = (0 . 955036 m / s)(1 . 11 s) = 1 . 06009 m . Question 3, chap 4, sect 6. part 3 of 3 10 points What is the average velocity of the skater relative to the water? Correct answer: 0 . 719 m / s (tolerance ± 1 %). Explanation: The time to go upstream is 1 . 11 s. The time to drift back downstream is t 1 = l v wg = . 262 m . 719 m / s = 0 . 364395 s , so the total time is t tot = t + t 1 = 1 . 11 s + 0 . 364395 s = 1 . 4744 s . Therefore, ¯ v = d sw t tot = 1 . 06009 m 1 . 4744 s = 0 . 719 m / s . Question 4, chap 4, sect 4. part 1 of 3 10 points A cannon sends a projectile towards a tar- get a distance 1220 m away. The initial ve- locity makes an angle 30 ◦ with the horizontal. The target is hit. The acceleration of gravity is 9 . 8 m / s 2 . What is the magnitude of the initial veloc- ity? Correct answer: 117 . 497 m / s (tolerance ± 1 %). Explanation: For a projectile, we have d = v 2 sin 2 θ g v = radicalBigg d g sin(2 θ ) = radicalBigg (1220 m) (9 . 8 m / s 2 ) sin(2 × 30 ◦ ) = 117 . 497 m / s . oldhomewk 09 – YOO, HEE – Due: Feb 5 2008, 4:00 am 2 Question 5, chap 4, sect 4. part 2 of 3 10 points How high is the highest point of the trajec- tory? Correct answer: 176 . 092 m (tolerance ± 1 %). Explanation: h = v 2 sin 2 θ 2 g = (117 . 497 m / s) 2 sin 2 30 ◦ 2 (9 . 8 m / s 2 ) = 176 . 092 m . Question 6, chap 4, sect 4. part 3 of 3 10 points How long does it take for the projectile to reach the target? (Assume no friction) Correct answer: 11 . 9895 s (tolerance ± 1 %)....
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This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Oldhw9_s - oldhomewk 09 – YOO HEE – Due Feb 5 2008 4:00 am 1 Question 1 chap 4 sect 6 part 1 of 3 10 points A skater(water spider maintains an

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