Question

Aroldis Chapman posses a 47.7 m/s fastball. If he threw a baseball straight up into the air, how long will the baseball stay in the air?

Answer 1

The time the baseball will stay in the air is twice its raise time (see http://www.college-physics.com/book/mechanics/vertically-launched-projectiles/). Thus, obtain:

$$t = \dfrac{2v_0}{g}$$

where $v_2 = 47.7m/s$ is the initial speed, and $g = 9.8m/s^2$ is the gravitational acceleration. Thus, obtain:

$$t = \dfrac{2\cdot 47.7m/s}{9.8m/s^2} = 9.7s$$

Answer. 9.7s.