SOLUTION: The height in feet of an object thrown upward is given by the equation h=80t-16t^2, where is the height of the object after 1 seconds. after how many seconds does the object reach

Question 470865: The height in feet of an object thrown upward is given by the equation h=80t-16t^2, where is the height of the object after 1 seconds. after how many seconds does the object reach its maximum height?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
The height in feet of an object thrown upward is given by the equation h=80t-16t^2, where is the height of the object after 1 seconds.
.
set t to 1 and solve for h:
h=80t-16t^2
h=80(1)-16(1)^2
h=80-16
h=64 feet
.
after how many seconds does the object reach its maximum height?
find the "axis of symmetry":
t = -b/(2a)
t = -80/(2(-16))
t = -80/(-32)
t = 2.5 seconds

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