a basketball player can jump 1.5m off the hardwood floor with what upward velocity did he leave the floor
m g h = 1/2 m v^2
v = √(2 g h)
To find the upward velocity with which the basketball player left the floor, you can use the equations of motion and the concept of projectile motion. The equation you can use is:
v² = u² + 2as,
where:
- v is the final velocity (which is 0 when the player reaches the maximum height),
- u is the initial velocity (what we want to find),
- a is the acceleration (which is equal to -9.8 m/s² on Earth due to gravity), and
- s is the displacement (the vertical height the player jumps, which is 1.5 m).
Rearranging the equation, you get:
u = √(v² - 2as),
Substituting in the values:
u = √(0 - 2 * (-9.8) * 1.5),
u = √(29.4),
u ≈ 5.43 m/s.
Therefore, the upward velocity with which the basketball player left the floor is approximately 5.43 m/s.