Question

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

Answer 1

well they tell you that v = 50

So what does the equation look like now?

Answer 2

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 \(t=0\)?

Answer 3

put in \(t=0\)

Answer 4

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

Answer 5

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

Answer 6

so what is the function?

Answer 7

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?

Answer 8

Now we can start to actually answer the question.

Answer 9

Ok part A wants to know \(t\) when the thing hits the ground. What is the height when something hits the ground?

Answer 10

ok so we need to solve \[-16t^2+50t+90=0\]

Can you do that?

Answer 11

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?

Answer 12

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

Answer 13

what do you get?

Answer 14

what is \(a\) and \(b\) in your equation?

\(ax^2+bx+c\)?

Answer 15

nono

\(-16x^2+50x+90\)
\( \ \ \ ax^2+bx+c\)

What is \(a\)?

Answer 16

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

Answer 17

2*-16 = 100?

Answer 18

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

Answer 19

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.

Answer 20

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

Answer 21

err 25/16