A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2 and find your answer to the nearest 0.001 s.

From the equations of linear motion,

v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,

Thus, v² = u² - 2gs, but v=0

hence, u² = 2gs

= 2*9.81*0.43

= 8.4366

u = √8.4366

=2.905 m/s

Hence the initial velocity is 2.905 m/s

Then using the equation v = u + gt.

Therefore, v = u - gt. (-g because the player is jumping against the gravity)

but, v = 0

Thus, u = gt

Hence, t = u/g

= 2.905/9.81

= 0.296 seconds