Aball was kicked into the air from a balcony 20 feet above the ground, and the ball’s height above the ground, in feet, t seconds after the ball was kicked was 2h(t) = 20 − 16t + 32t. what was the maximum height, in feet, of the ball above the ground after it was kicked?

2h(t) = 20 - 16t + 32t

Not only is there something wrong with the way you've written the equation,
there's no way it could be true even after you fix the little mistake.

First let's divide each side by 2 :

h(t) = 10 - 8t + 16t

This is the equation of the height of a ball that starts from 10-ft above the ground, not 20, begins with 8 ft/sec of downward speed as soon as it's kicked, and what to do with that mysterious ' 16t ' at the end ?  Well, if it were ' 16t-squared ', it very well could reflect the acceleration of gravity.

BUT ... since the '10' is positive, we know that the upward direction is the
positive direction for this problem, and then the sign of the acceleration term
can't be positive.  That would mean that the ball is accelerating upward at the
rate of 32 ft/sec every second, and if we just wait a few minutes, that ball is on
its way to the moon !

-- If we accept the equation exactly as it's written in the question, then the ball
is kicked from an initial height of 10-ft, it has an upward speed of +8 ft/sec forever,
it never sinks lower than the initial 10-ft, and it has no maximum height.

-- If we make the last term ' 16t² ', then the ball is kicked from an initial height
of 10-ft, the kicker aims down and gives it an initial speed of 8 ft/sec DOWNward,
but it has an upward acceleration of 32 ft/sec every second, The lowest it ever gets
is 1-ft below the balcony, at exactly 0.25 second after the kick, then begins rising,
faster and faster.  In this case also, the ball is headed for the moon, and has no
maximum height.

The question needs some serious work.

The answer is 109 times per second

Idon’t know what the answer is i wish i could