Question is to big for here its down there....?

Question

In this question, use the information s(t)=-16t2+v0t+s0. You may also
use your calculator on this question.
We throw a rock into the air with initial velocity of 50 ft/sec, and
initial position of 6 ft. Find how high the rock goes before coming back down. To answer this question, fill in the blanks in the following sentence, where the first blank gives this height, and the second blank gives how many seconds into the flight of the rock it reaches this maximum height.

The rock reaches its highest position of _______ feet above the ground
after _______ seconds of flight

Answer 1

s(t) = -16t² + 50t + 6

Put it in parabola form:

s(t) = -16(t² - 25/8 t) + 6

s(t) = -16(t² - 25/8t + 625/256) + 6 + 625/16

s(t) = -16(t - 25/16)² + 721/16

So the maximum is reached at:

t = 25/16 = 1.5625 seconds

And the maximum height achieved is:

s(25/16) = 721/16 = 45.0625 feet

If I didn't mess up.

Answer 2

s(t) = -16t^2 + 50t +6

To compute the max of s(t) take the first derivative,
s'(t) = -32t+50, set it to 0, and solve for t.

t = 50/32 = 25/16

max = s(25/16) = (25/16)^2 + 50(25/16) + 6