Can you help me solve this question?

Question

A ball is thrown vertically upward. After t seconds, its height h  (in feet) is given by the function h(t)=80t-16t². 

What is the maximum height that the ball will reach?

Do not round your answer

Answer 1

H(t) = -16t^2 + 80t ---> a= -16, b = 80, c=0

-b/[2a] = -80/[2(-16)] = -80/-32 = 40/16 = 20/8 = 10/4 = 5/2

-16(5/2)^2 + 80(5/2) = -16(25/4) + 80(5/2) = -100 + 200 = 100

(4ac - b^2)/(4a) = (4*-16*0 - 80^2)/(4*-16) =

= -6400/-64

= 100

max is 100 feet