SOLUTION: The height in feet of a projectile with an initial velocity of 96 feet per second and an initial height of 256 feet is a function of time in seconds given by h(t) = -16t^2 + 96t

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Question 1080958: The height in feet of a projectile with an initial velocity of 96 feet per second and an initial height of 256 feet is a function of time in seconds given by
h(t) = -16t^2 + 96t + 256.
(a) Find the maximum height of the projectile.
_______ft
(b) Find the time t when the projectile achieves its maximum height.
t =______sec
(c) Find the time t when the projectile has a height of 0 feet.
t =______sec
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
The t value is -b/2a=-96/-32 or 3 seconds (b)
The height will be -144+288+256=400 feet
-16t^2+96t+256=0
divide by -16
t^2-6t-16=0
(t-8)(t+2)=0
t=8 seconds, only positive root.
graph%28300%2C300%2C-10%2C10%2C-30%2C500%2C-16x%5E2%2B96x%2B256%29

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Look into the lessons
    - Problem on a projectile moving vertically up and down
    - Problem on a toy rocket launched vertically up from a tall platform
in this site.

From these lessons, learn on how to solve your problem in all details.


Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".