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

4410 m

It creates dogx^2 and then flip your answer and multiply by 2