During a baseball game, a batter hits a popup

Question

to a fielder 80 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 5.5 s, how
high does it rise?
Answer in units of m.

Answer 1

I do not care how far it went to the fielder.
I am only interested in the vertical problem.
it spent 5.5/2 seconds rising

v = Vi - g t
at top, v = 0
0 = Vi - 9.8 (5.5/2)
so
Vi = 27 m/s

so if you throw up at 27 m/s, how high?
two ways:
energy:
(1/2) m Vi^2 = m g h
so
h = Vi^2/(2g) = 27^2/(2*9.8) = 37 meters
or using integration
h = Vi t - 4.9 t^2
h = 27 (2.75) - 4.9 (2.75)^2
= 74 - 37
= 37 meters again