Question 1154333
h(t) = -16t^2 + 80t + 9.

compare with equation {{{ax^2+bx+c=0}}}

a=-16, b=80 c =9

time to go to height h  = {{{-b/2a}}}

=(80/32)= 2.5 s


time to go maximum height  is  t =  =  =  = 2.5 seconds.

 

The maximum height you can get  by substituting this value  t= 3.5 in the formula 
h(t) = -16t^2 + 80t + 9

h(2.5) = -16*2.5^2 + 80*2.5 + 9.

h =109m