Question

a basketball player jumps straight up for a ball. to do this, he lowers his body 0.34 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 0.77 m above the floor.

Answer 1

The velocity at which he leaves the floor is 4.25 m/s.

The kinematic expression is as follows:

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

when he reaches a height of 0.77 m above the floor the final velocity  = 0

acceleration due to gravity act opposite the initial direction of motion. So, -9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.77

- u² = 2 × - 9.81 × 0.77

multiply both sides by -1

u² = 2 × 9.81 × 0.77

u² = 15.107

u = √15.107

u = 4.2485762321

u = 4.25 m/s

The velocity at which he leaves the floor is 4.25 m/s.

To learn more about velocity, refer: brainly.com/question/18084516

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[NOTE : THIS QUESTION IS INCOMPLETE. THE COMPLEE QUESTION IS:

a basketball player jumps straight up for a ball. to do this, he lowers his body 0.34 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 0.77 m above the floor. Calculate his velocity (in m/s) when he leaves the floor.]