Question

PLEASE HELP I DONT UNDERSTAND AM I AM HORRIBLE AT GRAPHING....Recall the equation for projectiles is y = –16x2 + vx + s, where y is the object’s current height (in feet) at any time after it is projected, x is the time (in seconds) the object has been in the air, v is the object’s starting upward velocity, and s is the object’s starting height.

Answer 1

Do you need to graph it? Or something else?

Answer 2

use graphing technology to complete this activity since many of the solutions will most likely involve irrational numbers.... I REALLY DONT GET IT AT ALL. i really need help, do you get it ?

Answer 3

What is one of the questions that you are working on that uses this equation?

Answer 4

That may give you the correct answer, but it will make you worse off in the long run.

Answer 5

yea i want to learn how to do it not just get the answer, lol i do take my eoc's and want a future !

Answer 6

@aacehm im going to ask my teacher about that one, i have others would u like to help ?

Answer 7

Sure, go ahead

Answer 8

Motor a speed is 190 ft per sec, motor b is 210 ft per secs. and motor c is 230 ft per sec. You have various motor speed options but need to choose the one that most closely meets the client’s height request of 690 feet. useing the projctile formula y=-16x^2+vx+s w/ starting height of 0(s=0)

Answer 9

Okay, so how do you think you should start?

Answer 10

and then it says Substitute the data provided for speed (v) and starting heights (s) into their proper positions in the formula
y = –16x2 + vx + s and write the equation corresponding with each motor in the space provided below.
Then identify the a, b, and c values for each where indicated. (6 points total, 2 points each)

The equation for motor A is:
a =
b =
c =

The equation for motor B is:
a =
b =
c =

The equation for motor C is:
a =
b =
c =
what do i do ?

Answer 11

i assume i start with the formula they provide

Answer 12

okay, for the a values, they are all the same. Do you remember the form for quadratic equations?

Answer 13

ax2 + bx + c = 0 ???

Answer 14

right. So according to the equation -16x^2 + vx + s, what is a?

Answer 15

-16 ?

Answer 16

right! Okay, so now on to c, because that's easy too. What does c correspond to in the equation -16x^2+vx+s?

Answer 17

vx ?

Answer 18

nonoo i mean s

Answer 19

yes, s. And in the problem, the starting height is zero, so what is c for each equation?

Answer 20

i think 0

Answer 21

yes, good job! Okay, so now where do you think the speed of the motor goes in the equation?

Answer 22

um vx?

Answer 23

im not sure on that one

Answer 24

well, in this case you need to keep x, so you substitute the motor speed for v. Lastly, where does the 690 go?

Answer 25

after the = sign since thats the guys expetations ?

Answer 26

no, the desired height is 690, so you substitute 690 for y. Now, putting everything together, what is the equation for motor a?

Answer 27

-16x^2+190x+0=690 ?

Answer 28

Yes! Good job! Alright, so do you know how to solve quadratic equations? How would you solve this one?

Answer 29

im really not sureee....:// i feel dumb

Answer 30

Okay, here we go.\[-16x^{2}+190x=690\]\[-16x^{2}+190x-690=0\]

Answer 31

Now, what can you factor out of this equation. In other words, what is the GCF of 16, 190, and 690?

Answer 32

2 ?

Answer 33

Yeah
so factoring out a -2 would be good, since it's easier when a is positive. then we get
\[-2(8x^{2}-95x+345)=0\] So now we have to factor. Do you remember how to factor? What two numbers multiply to 345 and add to -95? This is kind of tricky

Answer 34

actually nevermind don't do that

Answer 35

alright

Answer 36

just use the quadratic formula, do you remember that?

Answer 37

ax2 + bx + c = 0

Answer 38

no, the one
\[\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]

Answer 39

oh my,i dont recell how to do that one, but yes im familiar with it

Answer 40

actually, you can use a graphing calculator right?
graph all three of the equations\[-16x^{2}+190x-690\]\[-16x^{2}+210x-690\]\[-16x^{2}+230x-690\]

Answer 41

I dont have one, and the one downloaded on my labtop isn't working for some reason. to turn my work in i have to draw it on fraph paper and scan it but im horrible with graphs

Answer 42

ugh. this is pretty tough.

Answer 43

you really need a graphing calculator for it

Answer 44

im sorry, you dont have to help me if you dont want. but i thank you for how far youve gotten me so far. appreciate it much

Answer 45

wait hold on ima see if they have a free one online

Answer 46

https://www.desmos.com/calculator

Answer 47

so if you graph the following three functions
\[-16x^{2}+190x\]\[-16x^{2}+210x\]\[-16x^{2}+230x\]

Answer 48

that is what the rockets' paths would look like if they were actually fired

Answer 49

so which one is closest to 690 at the peak of its arc?

Answer 50

it keeps saying error, can you maybe post a pic of them graphed ?

Answer 51

wait im not suppost to put the -690 ?

Answer 52

well, if you put the 690 you would just have to see which is closest to 0 at its peak instead of 690, whichever one you prefer

Answer 53

its the motor a ?

Answer 54

no, if you look the peak of motor b is actually closest

Answer 55

oh

Answer 56

so that's the answer, motor b

Answer 57

oh alright thank you i appreiate your help !

Answer 58

ONE MORE THING, how woulldi find the vertex coordinates by looking at the graphs @aacehm

Answer 59

no problem
um. you just do -b/2a and f(-b/2a)

Answer 60

Thank you @aacehm can you help me wit one more ?