Question

A basketball player of mass m=86.4 kg jumps straight up for a ball. To do this, he lowers his body d₁=0.36 m, and then he accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a speed sufficient to carry him d₂=0.47 m above the floor.

Write an expression for the acceleration, a, of the basketball player during the leg-straightening.

Answer 1

To calculate the basketball player's velocity when he leaves the floor, we use the conservation of energy principle. The acceleration of the player while straightening his legs can be found using the equation v = u + at. The force exerted by the player on the floor can be calculated using Newton's second law through which we get F ≈ 581.09 N

Step-by-step explanation:

To calculate the basketball player's velocity when he leaves the floor, we can use the conservation of energy principle. The total mechanical energy before the player jumps is equal to the total mechanical energy after the player jumps. At the start, the player has potential energy while crouching down, and at the end, the player has potential energy and kinetic energy. Therefore, the equation is:

mgh + 0.5mv^2 = mgh + 0.5mv^2

Simplifying, we get:

mgh = 0.5mv^2

Substituting the given values:

(86.4 kg)(9.8 m/s²)(0.36 m) = 0.5(86.4 kg)v^2

Solving for v:

v ≈ 2.42 m/s

To calculate the acceleration of the basketball player while he is straightening his legs, we can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity (which is 0 since the player starts from rest), a is the acceleration, and t is the time taken to accelerate. Rearranging the equation, we get:

a = (v - u) / t

Substituting the given values:

a = (2.42 m/s - 0 m/s) / 0.36 m

Simplifying, we get:

a ≈ 6.72 m/s²

Finally, to calculate the force the player exerts on the floor, we can use Newton's second law:

F = ma

Substituting the given values:

F = (86.4 kg)(6.72 m/s²)

Simplifying, we get:

F ≈ 581.09 N