#### Answer

1s. 1s.

#### Work Step by Step

a. To find the hang time, concentrate on the vertical component of velocity and the vertical displacement.
As far as vertical motion is concerned, in the second half of the jump, the person started at rest, and dropped 1.25 m. Find the time needed to fall that distance.
$$1.25 m = \frac{1}{2}g t^{2} = 5 \frac{m}{s^{2}}t^{2}$$
$$ t = 0.5 s$$
This is the time to drop downward, and we double this to find the hang time of 1 s.
This is discussed on page 50.
b. Hang time depends only on the vertical component of motion.