Re: come on
toasty said:
ive had this up 4 quite awhile and i got one no helpful respose. Come on. everyother post has gotten a repsonse that helped them except mine. geez.
Try to be patient. Most of the people here who help know a lot, but not everyone knows how to do everything, or maybe they spent so much time on another post, they decided to take a break. I've been there.
The height of the second equation is found the same way. the constant added on the end has no affect, it is still \(\displaystyle \frac{-b}{2a}\). When you take calculus or physics in the future you will learn why.
The time that it takes to get there can be found by using the quadratic equation (or by factoring if possible).
For example, if you are given: h = -2t^2 +4t - 1, the maximum height is:
\(\displaystyle \frac{-4}{2*(-2)} = \frac{-4}{-4} = 1\)
The time that it takes to get to its maximum height of 1 can be found by putting 1 in for h:
\(\displaystyle 1=-2t^2+4t-1\)
\(\displaystyle 2t^2-4t+2=0\)
You can factor this, or use the quadratic equation. Since it is factorable,
\(\displaystyle (2t-2)(t-1) = 0\).
From here we get t=1.
Get the idea?