Integrated Concepts A 105-kg basketball player crouches | StudySoup

Textbook Solutions for College Physics

Chapter 7 Problem 69

Question

Problem 69PE

Integrated Concepts A 105-kg basketball player crouches down 0.400 m while waiting to jump. After exerting a force on the floor through this 0.400 m, his feet leave the floor and his center of gravity rises 0.950 m above its normal standing erect position. (a) Using energy considerations, calculate his velocity when he leaves the floor. (b) What average force did he exert on the floor? (Do not neglect the force to support his weight as well as that to accelerate him.) (c) What was his power output during the acceleration phase?

Solution

Part (a)

Step 1 of 8:

A basketball player crouches down and jumps. During the jump the center of gravity of the player changes. When the player jumps, the center of gravity changes to a new position. We need to calculate the velocity of the player when he jumps using energy considerations.

The mass of the player

The acceleration due to gravity

The center of gravity

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full solution

Title College Physics  1 
Author Paul Peter Urone, Roger Hinrichs
ISBN 9781938168000

Integrated Concepts A 105-kg basketball player crouches

Chapter 7 textbook questions

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