Question 140798: The path of a fallen object is given by the function s= -16t^2+v0t+s0, where v0 represents the initial velocity in ft/sec and s0 represents the initial height. The variable t is time in seconds and s is the height of the object in feet.
If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40 foot building,how high is the rock after 0.5 seconds? After how many seconds will the rock reach maximum height?
Ive already answered the first question, i think it is 52 feet. Is this correct?
How do i set up my equation to answer the second question? (step by step please)
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! PART ONE
Yes, you are correct.
52 feet is the correct height.
PART TWO
I'm not sure how advanced you are in mathematics to solve the next part.
If you have taken calculus, you take the derivative of the function and set it equal to zero.
If you are using algebra, you can recognize that the equation is the equation of a parabola, find the axis of symmetry, and calculate the maximum there.
Finally you could plot points and try to find the maximum that way.
We can do all three.
1.
The maximum occurs at t=1 second.
2.
The axis of symmetry for the parabola
 occurs at 
In your case,
a=-16
b=32
The axis of symmetry is then
Just as above, s=56 ft at t=1 sec.
3. Start at t=0 and solve for s.
(0.0,40)
(0.5,52)
(1.0,56)
(1.5,52)
(2.0,40)
You could also take some points aroung t=1 to prove to yourself that the maximum does occur at t=1.
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