Question

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.260 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 0.920 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor.

Answer 1

Answer:

u = 4.25 m/s

Explanation:

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 0.920 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) × 0.920

- u² = 2 × - 9.81 × 0.920

multiply both sides by -1

u² = 2 × 9.81 × 0.920

u² = 18.0504

u = √18.0504

u = 4.2485762321

u = 4.25 m/s