During a baseball game the batter hits a high pop-up directly over home plate. If the ball stays in the air sixty seconds, how high did the ball go

Define

h = height (m) at time t (s).

Because the ball stays in the air for 60 s, the ball reaches a maximum height after 30 s, when the upward velocity is zero.

Note that

g = 9.8 m/s², the acceleration due to gravity.

Let v = the vertical launch velocity. Then

v - (9.8 m/s²) * (30 s) = 0

v = 294 m/s

Also, the maximum height, H, is

H = (294 m/s) * (30 s) - (1/2) * (9.8 m/s²) * (30 s) ²

= 4410 m

Answer: 4410 m