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
