If michael jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time

The formula for the vertical height, h, attained with a vertical take off speed of u is

u² - 2gh = 0

where

g = 9.8 m/s², acceleration due to gravity.

Because h = 1.29 m, therefore

u² = 2*9.8*1.39 = 27.244

u = 5.2196 m/s

Also, the time of flight, t, required when the vertical take-off speed is u is given by

ut - 0.5gt² = 0.

Therefore

5.2196t - 0.5*9.8*t² = 0

5.2196t - 4.9t² = 0

t (5.2196 - 4.9t) = 0

t = 0, or t = 5.2196/4.9 = 1.0652 s

t = 0 corresponds to take-off.

t = 1.0652 s corresponds to landing.

The hang time is 1.0652 s.

Answer:

Take-off speed = 5.22 m/s (nearest hundredth)

Hang time = 1.065 s (nearest thousandth)