anonymous
  • anonymous
Frank kicks a soccer ball off the ground and in the air with an initial velocity of 30 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches? 14.1 feet 13.7 feet 13.2 feet 15.2 feet
Mathematics
jamiebookeater
  • jamiebookeater
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zepp
  • zepp
We are looking for the highest point of a facing down parabola, therefore the vertex of the parabola.
zepp
  • zepp
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anonymous
  • anonymous
Do your thing, Zeppy.

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anonymous
  • anonymous
so this means??
anonymous
  • anonymous
lol i should have read the chapter :/
zepp
  • zepp
|dw:1341586632841:dw|
zepp
  • zepp
Oh
anonymous
  • anonymous
GAHAHAHA YEEEESSSS
anonymous
  • anonymous
lol so whats the answer, nice drawing
anonymous
  • anonymous
The eeeeye, it sees everything!
anonymous
  • anonymous
lol i moved my webcam away from me haha
zepp
  • zepp
H(t) = −16t2 + vt + s Represents the equation of your parabola, we are looking for the vertex, the x-value of the vertex could be found by using the formula \[\frac{-b}{2a}\]
anonymous
  • anonymous
so one of these 14.1 feet 13.7 feet 13.2 feet 15.2 feet
zepp
  • zepp
Now, 30 feet per second. v=30 H(t) = −16t2 + 30t + s
zepp
  • zepp
Use the formula I wrote above :)
anonymous
  • anonymous
dont know how to solve that :/ i just need the answer no offence
anonymous
  • anonymous
ok il guess thanks for the help (in a good way)
zepp
  • zepp
A quadratic equation is \(ax^2+bx+c\) Where the x of the vertex is at \(\large \frac{-b}{2a}\) In our equation H(t) = −16t2 + 30t + s b is 30 and a is -16 \[\frac{-30}{2(-16)}=\frac{-30}{-32}=\frac{30}{32}\]
zepp
  • zepp
What you have to do next it to replace this x-value back into the equation to find the height (y-value) And sorry I'm at work so I was afk :(
anonymous
  • anonymous
what is it

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