PreCalc
posted by MUFFY Question has been fixed .
The height s of a ball, in feet, thrown with an initial velocity of 112 feet per second from an initial height of 20 feet is given as a function of time t (in seconds) by
s(t)= 16t^2+112t+20. What is the maximum height of the ball? At what time does the maximum height occur?
I know how to find the maximum height. You find the vertex which is 3.5 then plug it back into the original equation.
Now, when it says what time, I am not sure if it wants the vertex or should I use the quadratic to solve it?
Please help.

PreCalc 
MathMate
The maximum height is the vertex.
This would be an application of completing the squares to convert the function into the standard form:
s(t)=a(th)²+k
where t=h at the vertex (3.5) and
k=maximum height (216 above datum) 
PreCalc 
MUFFY Question has been fixed
Ok thanks.

PreCalc 
MathMate
You're welcome!
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