Calculate the hang time of an athlete who jumps a vertical distance of 0.75 meter

The time it takes going up will be equal the time returning:

s = s0 + v0t + (1/2) at^2

Taking the top of the jump as 0 and positive direction downward,

h = 0 + 0 + (1/2) gt^2

t = √ (2h/g)

Since the hang time will be twice the time to rise or fall, therefore:

T = 2t

T = 2√ (2*0.75/9.80665)

T ≈ 0.78 seconds