SOLUTION: If an object is propelled upward from a height of 96ft at an initial velocity of 80ft per second, then its height after t seconds is given by the equation h = -16t^2 + 80t + 96 whe

Question 328191: If an object is propelled upward from a height of 96ft at an initial velocity of 80ft per second, then its height after t seconds is given by the equation h = -16t^2 + 80t + 96 where h is in feet. After how many seconds will the object reach a height of 196ft?
I am not so good at word problems and am really confused! You have great tutors that I appreciate so very much. I hope that someone will lend me a hand this one has me stumped. Thank you in advance.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
an object is propelled upward from a height of 96ft at an initial velocity of 80ft per second,
its height after t seconds is given by the equation h = -16t^2 + 80t + 96 where h is in feet.
After how many seconds will the object reach a height of 196ft?
:
-16t^2 + 80t + 96 = h
Replace h with 196
-16t^2 + 80t + 96 = 196
-16t^2 + 80t + 96 - 196 = 0
-16t^2 + 80t - 100 = 0
Simplify & change the signs, divide by -4, (easier to factor)
4t^2 - 20x + 25 = 0
Factors to
(2t-5)(2t-5) = 0
2t = 5
t =
t = 2.5 seconds it will be 196 ft (this is also the max height)
:
Graphically; t on the x axis, h on the y axis

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