I need somebody to walk me through this, I understand the just of it. PLease Help! Medal and Fan!

- anonymous

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## More answers

- anonymous

@Donblue

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@nincompoop

- anonymous

What is the question? I might be able to help you.

- anonymous

The function H(t) = -16t2 + vt + s shows the height H(t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.
Part A: The projectile was launched from a height of 96 feet with an initial velocity of 80 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points)
Part B: What is the maximum height that the projectile will reach? Show your work. (2 points)
Part C: Another object moves in the air along the path of g(t) = 31 + 32.2t where g(t) is the height, in feet, of the object from the ground at time t seconds.
Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points)
Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know? (2 points)

- anonymous

Is this FLVS?

- anonymous

Yep!

- anonymous

Algebra?

- anonymous

Algebra 1 semester 2, just this and im done the whole thing

- anonymous

did you plug in the numbers to create the equation for A?

- anonymous

i tried to but didnt quite understand

- anonymous

Ok I just finished it yesterday. Maybe I can see if I had it too

- anonymous

this is what i got
H(t) = -16(96)^2 + 80(96) + 96

- anonymous

\(H(t)=-16t^2+vt+s\) is the basic equation. You need to replace the v with the velocity they gave you, and replace the s with the initial height. t is the variable, so it stays the same.

- anonymous

all you need to do is change v to 80 and s to 96

- anonymous

so the velocity is 80 fps
and the height is 96
but whats the time?

- anonymous

?

- anonymous

what about t?

- anonymous

time is the independent variable. the projectile has a different height for different values of time. It's 96 in the beginning and then goes up and comes back down

- anonymous

i actually do need time to type, you know?

- anonymous

sorry, okay s it would be H(t) = -16t^2 + 80t + 96?

- anonymous

yes

- anonymous

Sweet! now could you help with b?

- anonymous

would you solve for t?

- anonymous

Yes, for B, the height's going to be the y-coordinate of the vertex using the equation from a.

- anonymous

\[t=-\frac{ b }{ 2a }\]
Look familiar?

- anonymous

then once you have that t, plug it in to get the height

- anonymous

Yes, okay soi got 5/2?

- anonymous

that's the right time. now put it in for t in the equation from A

- anonymous

\[-16(5/2)^2+80(5/2)+96\]

- anonymous

oh okay, and then solve?

- anonymous

yes

- anonymous

i got 196?

- anonymous

yes that's right

- anonymous

C?

- anonymous

for c, create a table for the values 0 through 5 and plug them into both equations

- anonymous

You're looking for the t where you get the same (or pretty close) H for both equations

- anonymous

okay let me see

- anonymous

is it possible to solve algebraicly without a table?

- anonymous

yes, but you probably haven't done it in algebra I. |dw:1436390010929:dw|

- anonymous

you'd set the equations equal to each other then solve for t
-16t^2 + 80t + 96 = 31+32.2t

- anonymous

it's a standard quadratic equation, so maybe you've done that

- anonymous

my bad didnt mean to type that

- anonymous

both aanswers are rounded but i got t = -1, t = 4

- anonymous

you want to go with 4 because time has to be positive

- anonymous

okay, so thats the answer for c?

- anonymous

yes

- anonymous

for D, 4 seconds is after the max height at 2.5 seconds, so it's got to be on the way down

- anonymous

okay, thank you so much! that really helped a lot :)

- anonymous

you're welcome

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