Question

Velocity sufficient to carry him 0.920 m above the floor. (a) Calculate his velocity (in m/s ) when he leaves the floor. (Enter a number.) ]m/s (b) Calculate his acceleration (in m/s²) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.340 m. (Enter a number.) m/s²

(c) Calrulato tho fnrce (in N) he exerts on the floor to do this, given that his mass is 104 kg. (Enter a number.)

Answer 1

To calculate the basketball player's velocity when he leaves the floor, we can use an equation of motion. For part (b), we can use a similar equation of motion to calculate the player's acceleration while straightening his legs. For part (c), we can calculate the force exerted on the floor using Newton's second law.

Step-by-step explanation:

To calculate the basketball player's velocity when he leaves the floor, we can use the equation of motion:

Vf^2 = Vi^2 + 2ad

Where Vf is the final velocity, Vi is the initial velocity (which is 0, as the player starts from rest), a is the acceleration, and d is the distance traveled. Plugging in the given values for the distance (0.900 m) and the acceleration (9.8 m/s^2, assuming acceleration due to gravity), we can solve for Vf to find the player's velocity when leaving the floor.

For part (b), we can use a similar equation of motion to calculate the player's acceleration while straightening his legs:

Vf^2 = Vi^2 + 2ad

Where Vf is the final velocity (the velocity found in part (a)), Vi is the initial velocity (0), a is the acceleration, and d is the distance traveled (0.300 m). Solving for a will give us the player's acceleration.

Lastly, for part (c), we can calculate the force exerted on the floor by using Newton's second law: F = m*a. Plugging in the given mass (104 kg) and the acceleration found in part (b), we can calculate the force exerted on the floor.