Abasketball 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