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well they tell you that v = 50
So what does the equation look like now?

put in \(t=0\)

nono, so if we let \(t=0\) into the function we get \(s\) correct?

right, and they told us that the intial height is 90, so s=90. make sense?

so what is the function?

Now we can start to actually answer the question.

ok so we need to solve \[-16t^2+50t+90=0\]
Can you do that?

the \(x\) coordinate of the vertex of the function \(ax^2+bx+c\) is \(x=\dfrac{-b}{2a}\).

what do you get?

what is \(a\) and \(b\) in your equation?
\(ax^2+bx+c\)?

nono
\(-16x^2+50x+90\)
\( \ \ \ ax^2+bx+c\)
What is \(a\)?

correct, ok so what is \(\dfrac{-b}{2a}\)?

2*-16 = 100?

correct, now plug that into the equation to get the height.

so plug in -25/16 into the function and that will give you the max height.

err 25/16