Question

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

Answer 1

The velocity of the basketball player when leaving the floor is 4.20 m/s. The acceleration while straightening legs is 29.4 m/s^2. The force exerted on the floor is 4,314 N.

Step-by-step explanation:

To solve these problems, we can use the principles of kinematics and dynamics in physics. First, we need to calculate the velocity with which the basketball player leaves the ground to reach a height of 0.900 m.

Part A: Using the kinematic equation for constant acceleration under gravity (assuming upward is positive):
v2 = u2 + 2as where v is the final velocity (0 m/s at the peak of the jump), u is the initial velocity, a is the acceleration due to gravity (-9.81 m/s2), and s is the displacement (0.900 m). Solving for u gives:

u = sqrt(2 * 9.81 m/s2 * 0.900 m) = 4.20 m/s.

Part B: The equation v2 = u2 + 2as can also be used to find acceleration while the player is straightening legs, with u = 0 m/s and s = 0.300 m. Solving for a gives:

a = v2 / (2s) = (4.20 m/s)2 / (2 * 0.300 m) = 29.4 m/s2.

Part C: To find the force, we'll use Newton's second law: F = ma. The total force is the force needed to accelerate the player's mass plus the force required to overcome gravity (weight).

F = (110 kg)(29.4 m/s2) + (110 kg)(9.81 m/s2) = 4,314 N (upward)

Answer 2

a) Vo = 4.24m/s

b) a = 30m/s^2

c) F = 4400N

Step-by-step explanation:

For part a) we study the movement after he leaves the ground. From that we know:

`Vf^2 = Vo^2 - 2*g*\Delta Y` Where Vf=0 and ΔY=0.9m

Solving for Vo:

`Vo = √(2*g*\Delta Y) = 4.24m/s`

For part b) we have to study the movement before he leaves the ground.

`Vf^2 = Vo^2+2*a*\Delta Y` Where Vo=0 and ΔY=0.3m

Solving for the acceleration:

`a=(Vf^2)/(2*\Delta Y)=30m/s^2`

For part c) we evaluate forces on the player

The force exerted by the floor on the player is the same as the one exerted by the player on the floor:

Fe - m*g = m*a Fe = m*(g+a) = 4400N