Question 106305: An object is shot upward from a height of 80 feet with an initial velocity of 64 feet per second. Its height (h) above the ground in feet after t seconds is given by the formula
h = - 16t2 + 64t + 80. At what times will the object be 128 feet above the ground?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! An object is shot upward from a height of 80 feet with an initial velocity of 64 feet per second. Its height (h) above the ground in feet after t seconds is given by the formula
h = - 16t2 + 64t + 80. At what times will the object be 128 feet above the ground?
:
use the formula and set the height (h) to 128 ft
-16t^2 + 64t + 80 = 128
:
-16t^2 + 64t + 80 - 128 = 0; subtract 128 from both sides
:
-16t^ + 64t - 48 = 0
:
Divide equation by -16, simplifies and changes the signs, easier to factor then
+t^2 - 4t + 3 = 0
Factors to:
(t - 3) (t - 1) = 0
Two solutions:
t = +1 sec, at 128 ft on the way up
and
t = +3 sec, at 128 ft on the way down
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