im new hi (moderators delete this please :) and my other question thanks)

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im new hi (moderators delete this please :) and my other question thanks)

Mathematics
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well they tell you that v = 50 So what does the equation look like now?
ok then they tell us that it starts off at 90 What do you get at the start of time? I.e. what do you get when \(t=0\)?
put in \(t=0\)

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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?
nono, ok before we can start to answer the question we had to figure out what \(v\) and \(s\) are. Now we have done that and we have our function is \[H(t) -16t^2+50t+90\] Are you with me?
Now we can start to actually answer the question.
Ok part A wants to know \(t\) when the thing hits the ground. What is the height when something hits the ground?
ok so we need to solve \[-16t^2+50t+90=0\] Can you do that?
I am not sure if that is correct, but you can check that. Ok notice that since we are dealing with time, that we would only take the positive answer. That is part A. For part B, do you know how to find the vertex?
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.
I am not responsible for you failing a class. I also highly doubt that getting help on one question will change you from a failing grade to passing.
so plug in -25/16 into the function and that will give you the max height.
err 25/16

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