You throw a beanbag in the air and catch it 2.2s later at the same place you threw it. how high did it go? what was its initial velocity?

The round trip took the beanbag 2.2 s which means that moving up and down will take it only 1.1s, respectively.

a. The distance covered by the beanbag is calculated through the equation,
                d = 0.5gt²
where d is distance, g is the acceleration due to gravity and t is time. Substituting the known values,
                d = 0.5(9.8)(1.1²) = 5.929 m

b. The initial velocity is calculated through the equation,
               v = gt
Substituting the known values,
               v = (9.8 m/s²)(1.1 s)
                 v = 10.78 m/s

Ithink it is a. but i'm not 100 percent sure.