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?

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

Step 2 of 8:

From conservation of energy

Total energy before jump = total energy after jump

Where

is the initial kinetic energy

is the initial potential energy which is zero

is the final kinetic energy at the height, it is also zero.

is the final potential energy at the height

Therefore the above equation becomes

Step 3 of 8:

Solving the above equation for velocity

Substituting the values

The velocity of the player when he jumps is 4.32 m/s

Part (b)

Step 4 of 8:

When the player jumps from the ground he exerts the force on the ground. There are two forces acting on the floor. One is the weight of the person and another one is the force required to make a jump. The player crouches himself a particular distance and jumps. A work is done by the player while crouching on the floor. We are going to calculate the average force exerted on the floor.

The mass of the player

The acceleration due to gravity

The center of gravity