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CP BIO A Standing Vertical Jump. Basketball player Darrell

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 39P Chapter 4

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 39P

CP BIO? A Standing Vertical Jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet left the floor.) Griffith weighed 890 N (200 lb). (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his aver-age acceleration (magnitude and direction) while he pushed against the floor? (c) Draw his free-body diagram. In terms of the forces on the diagram, what was the net force on him? Use Newton’s laws and the results of part (b) to calculate the average force he applied to the ground.

Step-by-Step Solution:

Solution 39P Step 1 : In this question, we need to find Speed with which Darrell griffith left the floir v Average acceleration ( magnitude and direction ) while he pushed against the floor Draw his body diagram and find net force on him Average force applied by player in ground Step 2 : Let us consider the data given Maximum height reached h = 1.2 m Mass of player m = 890 N or 200 lb Time taken for part of jump t = 0.300 s Step 3 : We shall find the initial velocity of the vertical jump using 1mv = mgh 2 Solving this for velocity we get v = 2gh Gravitational force g = 9.8 ms2 Substituting the values we get v = 2 × 9.8 m/s × 1.2 m v = 23.52 m /s 2 v = 4.849 m/s Hence we have obtained velocity as 4.849 m/s Step 4 : Let us find the acceleration of the player We have initial velocity as v = 0 i/s And final velocity as v = 4f489 m/s Time is given as t = 0.300s Thus we have use relation v a = t Substituting the values we get a = vfvi t 4.489 m/s 0 m/s a = 0.300 s a = 16.16 m/s 2 Hence we have acceleration as a = 16.16 m/s 2

Step 5 of 7

Chapter 4, Problem 39P is Solved
Step 6 of 7

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The answer to “CP BIO? A Standing Vertical Jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet left the floor.) Griffith weighed 890 N (200 lb). (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his aver-age acceleration (magnitude and direction) while he pushed against the floor? (c) Draw his free-body diagram. In terms of the forces on the diagram, what was the net force on him? Use Newton’s laws and the results of part (b) to calculate the average force he applied to the ground.” is broken down into a number of easy to follow steps, and 127 words. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 39P from 4 chapter was answered, more than 635 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 39P from chapter: 4 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: floor, jump, left, diagram, standing. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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