When no time has passed, t for time is 0.  We substitute 0 for t 

h(t) = −4.9t² + 24t + 8
h(0) = −4.9(0)² + 24(0) + 8
h(0) = 0 + 0 + 8
h(0) = 8 meters

and find that the ball is 8 meters above the ground when it leaves the
pitcher's hand.

When 1 second has passed, we substitute 1 for t 

h(t) = −4.9t² + 24t + 8
h(1) = −4.9(1)² + 24(1) + 8
h(1) = -4.9(1)² + 24 + 8
h(1) = 27.1 meters

and find that the ball is 27.1 meters above the ground.

When 2 seconds have passed, we substitute 2 for t 

h(t) = −4.9t² + 24t + 8
h(2) = −4.9(2)² + 24(2) + 8
h(2) = -19.6 + 48 + 8
h(2) = 36.4 meters

and find that the ball is 36.4 meters above the ground.

When 3 seconds have passed, we substitute 3 for t 

h(t) = −4.9t² + 24t + 8
h(3) = −4.9(3)² + 24(3) + 8
h(3) = -44.1 + 72 + 8
h(3) = 35.9 meters

and find that the ball is 35.9 meters above the ground. So the ball is not as
high after 3 seconds as it was after 2 seconds.  That means it's already
reached its maximum height as is coming down.  So it must have reached its
maximum height at more than 2 seconds and less than 3 seconds.



To find that time we use the vertex formula to find the number of seconds that
must have passed when it reached its maximum height, -b/(2a) = -(24)/[2(-4.9)]
= 2.448979592 seconds.  That's the answer.  We didn't need to do all that work
at the top.  We could have just jumped to the vertex formula.  But you said you
wanted a detailed version, so you'd understand what was going on with the
height of the ball after 1 second, 2 seconds, and 3 seconds.

To find out how high it reached, we substitute t=2.448979592 seconds and get
that its maximum height was 37.3877551 meters.

Edwin