HELP!!! I WILL DO ANYTHING

- anonymous

HELP!!! I WILL DO ANYTHING

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- 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 82 feet with an initial velocity of 60 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) = 10 + 63.8t 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

@dan815

- anonymous

@triciaal

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## More answers

- Plasmataco

A:0=-16t2+60t+82

- anonymous

YEAH, I have that part. but what do you mean by A;0 .. I have the numbers plugged in .. just solving is difficult

- anonymous

@jim_thompson5910

- Plasmataco

A: 0. Sry. Forgot the space

- Plasmataco

Just sayin for problem a.

- anonymous

either way what do you mean??

- anonymous

ohh

- anonymous

hello?

- Plasmataco

The equation for problem a is 0=-16t2+60t+82

- anonymous

I know that

- jim_thompson5910

part A is definitely 0 = -16t^2+60t+82. So you have the correct answer there for that part.
initial velocity = 60 ft/s
v = 60
initial height = 82 ft
s = 82

- jim_thompson5910

for part b, consider the equation y = -16x^2+60x+82
it is in the form y = ax^2 + bx + c

- anonymous

okay...
now what do i do?

- anonymous

yes

- jim_thompson5910

notice how a = -16, b = 60, c = 82
plug the values of a and b into h = -b/(2a) and tell me what you get for h

- anonymous

-1.875

- jim_thompson5910

it should be positive

- jim_thompson5910

you lost a sign somewhere

- anonymous

yea. it is . i accidentally put it there

- jim_thompson5910

now plug x = 1.875 into y = -16x^2+60x+82 to get the value of y
what is the value of y?

- anonymous

y=1094.5 ?

- jim_thompson5910

that's too big

- anonymous

,,,,, but i plugged the number in and I'm sure i solved it correctly..

- jim_thompson5910

-16x^2+60x+82
-16(1.875)^2+60(1.875)+82 ... replace every x with 1.875
now compute `-16(1.875)^2+60(1.875)+82`

- anonymous

yeah.. thats what i had. .let me retry

- anonymous

138.25??

- jim_thompson5910

much better

- anonymous

okay great.. next step?

- Plasmataco

For c, u make the equation in a equal 10 +63.8t
10+63.8t=-16t2+60t+82
In which you solve for t, then plug it in in one of the equations

- anonymous

HUH
^^^^

- jim_thompson5910

So when the time is t = 1.875 seconds, the height of the object is 138.25 ft which is the peak height
|dw:1439942710928:dw|

- jim_thompson5910

|dw:1439942749753:dw|

- anonymous

ohhhh okay

- Plasmataco

I thought u were done with b, sry for interuptingðŸ˜°

- anonymous

its ok...

- anonymous

now what?

- jim_thompson5910

what kind of calculator do you have?

- Plasmataco

So basically, 138.25 meters is the answer to b

- anonymous

Texas Instruments TI 30X

- Plasmataco

Right?

- jim_thompson5910

have you used a website called desmos before? It's a free graphing calculator

- jim_thompson5910

yes @Plasmataco

- anonymous

yes!!

- jim_thompson5910

ok we're going to use that to answer part c

- Plasmataco

Oh ok

- anonymous

ok what do i type in?

- Plasmataco

Scroll upsidoodles

- jim_thompson5910

type in f(x) = -16x^2 + 60x + 82 into the first box
and g(x) = 10 + 63.8x into the second box
let me know when you have that in and I'll move onto the next step

- Plasmataco

You want go and fx to be the same

- Plasmataco

Gx dang it autocorrect

- anonymous

okay done.. but i don't know how to put the f(x) and g(x) into the equation.. I don't see those as options.. like i can type the like equation after the = sign but nothing before

- jim_thompson5910

you should have this typed in so far

##### 1 Attachment

- anonymous

how did you get the f(x) there... just type?

- jim_thompson5910

yes

- anonymous

ok.. but no lines appear

- Plasmataco

If Gx=Fx, the -16t2+60t+82= 10+63.8t

- jim_thompson5910

your graph window may be off

- anonymous

i have a mac.. nothing is wrong with it.. so how is that possible?

- Plasmataco

Des ps is a lot quicker :3

- jim_thompson5910

try setting the y min to -20 and the y max to 150

- jim_thompson5910

click the wrench on the right side to adjust the graph window

- anonymous

i have lines now..

- jim_thompson5910

you should see something like this

##### 1 Attachment

- anonymous

yeah

- jim_thompson5910

now onto the table part
select the box under the g(x). So select the third box
then move up to the + button and add a table
change the \(\Large x_1\) to just x.
In the second column, replace \(\Large y_1\) with f(x). You'll see a list of numbers pop up
in the third column, type g(x) at the top of this column. More numbers will pop up

- jim_thompson5910

the goal is to have the numbers be equal. Which row does this? Or do we get pretty close at all?

- anonymous

they aren't appearing with numbers. i have the table and stuff.. but there aren't any numbers

- jim_thompson5910

you should see this

##### 1 Attachment

- anonymous

okay I got it now

- jim_thompson5910

do you see any rows where f(x) = g(x) or pretty close to it?

- anonymous

yea. row 2

- jim_thompson5910

yep, so at about t = 2 seconds, the two objects will be at the same height (approximately)

- anonymous

awesome! So is that part c??

- jim_thompson5910

yes

- anonymous

okay and Finally, Part D/

- jim_thompson5910

and it might be hard to see, but the projectile is falling down when they finally met up
here is a zoomed in look
|dw:1439944300088:dw|

- anonymous

ohhh. okay

- jim_thompson5910

if you click the intersection point between the two graphs, you'll see the point (2.006, 137.98) pop up. This is the more accurate approximate intersection point
So at approx t = 2.006 seconds, the two objects are at an approx height of 137.98 ft

- anonymous

Great :)

- anonymous

@jim_thompson5910 so if that is all.. I know your probably exhausted ...but could you help me with another problem I only need help on one part and its fairly simple???

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