SOLUTION: I would really appreciate it if I could get some help on this question, thanks. The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the init

Question 137456: I would really appreciate it if I could get some help on this question, thanks.
The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the initial velocity in ft/sec and so represents the initial height in feet. Also, s represents the height in feet of the object at any time,t,which is measured in seconds.
If a rock is thrown upward with the initial velocity of 40ft per second from the top of a 30ft building.
s= -16t^2+vot+so I figured out that much, then it asks after how many seconds will the graph reach maximum height? I just need help setting up the equation to solve for height....The question comes from my online algebra class at aiu. HELP PLEASE. thanks, alaska
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the maximum (or minimum) of a parabola is the vertex, which lies on the axis of symmetry

for a parabola of the form ax^2+bx+c, the equation for the axis of symmetry is x=-b/(2a)

so the time at maximum will be t=-40/(2(-16)) __ t=-40/(-32) __ t=1.25
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