anonymous
  • anonymous
HELP!!! I WILL DO ANYTHING
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • 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
  • anonymous
anonymous
  • anonymous

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Plasmataco
  • Plasmataco
A:0=-16t2+60t+82
anonymous
  • 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
  • anonymous
Plasmataco
  • Plasmataco
A: 0. Sry. Forgot the space
Plasmataco
  • Plasmataco
Just sayin for problem a.
anonymous
  • anonymous
either way what do you mean??
anonymous
  • anonymous
ohh
anonymous
  • anonymous
hello?
Plasmataco
  • Plasmataco
The equation for problem a is 0=-16t2+60t+82
anonymous
  • anonymous
I know that
jim_thompson5910
  • 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
  • jim_thompson5910
for part b, consider the equation y = -16x^2+60x+82 it is in the form y = ax^2 + bx + c
anonymous
  • anonymous
okay... now what do i do?
anonymous
  • anonymous
yes
jim_thompson5910
  • 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
  • anonymous
-1.875
jim_thompson5910
  • jim_thompson5910
it should be positive
jim_thompson5910
  • jim_thompson5910
you lost a sign somewhere
anonymous
  • anonymous
yea. it is . i accidentally put it there
jim_thompson5910
  • 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
  • anonymous
y=1094.5 ?
jim_thompson5910
  • jim_thompson5910
that's too big
anonymous
  • anonymous
,,,,, but i plugged the number in and I'm sure i solved it correctly..
jim_thompson5910
  • 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
  • anonymous
yeah.. thats what i had. .let me retry
anonymous
  • anonymous
138.25??
jim_thompson5910
  • jim_thompson5910
much better
anonymous
  • anonymous
okay great.. next step?
Plasmataco
  • 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
  • anonymous
HUH ^^^^
jim_thompson5910
  • 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
  • jim_thompson5910
|dw:1439942749753:dw|
anonymous
  • anonymous
ohhhh okay
Plasmataco
  • Plasmataco
I thought u were done with b, sry for interuptingšŸ˜°
anonymous
  • anonymous
its ok...
anonymous
  • anonymous
now what?
jim_thompson5910
  • jim_thompson5910
what kind of calculator do you have?
Plasmataco
  • Plasmataco
So basically, 138.25 meters is the answer to b
anonymous
  • anonymous
Texas Instruments TI 30X
Plasmataco
  • Plasmataco
Right?
jim_thompson5910
  • jim_thompson5910
have you used a website called desmos before? It's a free graphing calculator
jim_thompson5910
  • jim_thompson5910
anonymous
  • anonymous
yes!!
jim_thompson5910
  • jim_thompson5910
ok we're going to use that to answer part c
Plasmataco
  • Plasmataco
Oh ok
anonymous
  • anonymous
ok what do i type in?
Plasmataco
  • Plasmataco
Scroll upsidoodles
jim_thompson5910
  • 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
  • Plasmataco
You want go and fx to be the same
Plasmataco
  • Plasmataco
Gx dang it autocorrect
anonymous
  • 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
  • jim_thompson5910
you should have this typed in so far
anonymous
  • anonymous
how did you get the f(x) there... just type?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
ok.. but no lines appear
Plasmataco
  • Plasmataco
If Gx=Fx, the -16t2+60t+82= 10+63.8t
jim_thompson5910
  • jim_thompson5910
your graph window may be off
anonymous
  • anonymous
i have a mac.. nothing is wrong with it.. so how is that possible?
Plasmataco
  • Plasmataco
Des ps is a lot quicker :3
jim_thompson5910
  • jim_thompson5910
try setting the y min to -20 and the y max to 150
jim_thompson5910
  • jim_thompson5910
click the wrench on the right side to adjust the graph window
anonymous
  • anonymous
i have lines now..
jim_thompson5910
  • jim_thompson5910
you should see something like this
anonymous
  • anonymous
yeah
jim_thompson5910
  • 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
  • jim_thompson5910
the goal is to have the numbers be equal. Which row does this? Or do we get pretty close at all?
anonymous
  • anonymous
they aren't appearing with numbers. i have the table and stuff.. but there aren't any numbers
jim_thompson5910
  • jim_thompson5910
you should see this
anonymous
  • anonymous
okay I got it now
jim_thompson5910
  • jim_thompson5910
do you see any rows where f(x) = g(x) or pretty close to it?
anonymous
  • anonymous
yea. row 2
jim_thompson5910
  • jim_thompson5910
yep, so at about t = 2 seconds, the two objects will be at the same height (approximately)
anonymous
  • anonymous
awesome! So is that part c??
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
okay and Finally, Part D/
jim_thompson5910
  • 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
  • anonymous
ohhh. okay
jim_thompson5910
  • 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
  • anonymous
Great :)
anonymous
  • 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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