During a baseball game, a batter hits a pop-up to a fielder 92m away. The acceleration is 9.8 m/s^2. If the ball remains in the air for 5.8 seconds, how high does it rise?

Answer 1

Tf = Tr = 5.8/2 = 2.9 s.
Tr = Rise time.
Tf = Fall time.

h = 0.5g*Tf^2