How long tina, a ballerina, in the air when she leaps straight up with a speed of 1.8 m/s?

The acceleration of gravity on or near the surface of the Earth is 9.8 m/s².
Anything acted on only by gravity loses 9.8 m/s of upward speed, or gains
9.8 m/s of downward speed, every second.

Leaping straight upward at 1.8 m/s, Tina keeps rising until she runs out of
upward speed.  That happens in (1.8/9.8) = 0.1837 second after the leap.

After that, Finkel's First Law of Motion takes over:
"What goes up must come down."

The dropping part of the leap is symmetrical with the first.  Please don't
make me go through proving it.  Tina hits the floor at the same speed of
1.8 m/s with which she left it, and it takes the same amount of time to drop
from the peak to the floor as it took to rise from the floor to the peak.

So her total time out of contact with the floor is

                     2 x (0.1837 sec)  =  0.367 second  (rounded)

45 is the correct answer