Question

This portfolio is really confusing me, but I need help with these questions and I think I'm supposed to be working with the equation I made that is shown in the picture.

Answer 1

@Hero @pooja195

Answer 2

@jagr2713

Answer 3

@phi @zepdrix

Answer 4

can explain what the question is?
I see
h(t)= -16t^2 + 150 t + 18

what are you supposed to do with that ?

Answer 5

Want me to show you the whole thing?

Answer 6

yes, please

Answer 7

ok, and how much have you answered?

Answer 8

I have answered the first 2 questions along with question 3 part A, B, and C.

Answer 9

ok. For 3A, what height did you use for the building ?

Answer 10

I used 18.

Answer 11

ok. Just checking.

when will it land?
that is asking, when will the height be 0 (ground level)
so we have to find t when
\[ -16t^2 +150 t+18= 0 \]
I just did that and got t= 9.4935...
so that looks good

Answer 12

That is what I got as well.

Answer 13

I rounded it and said it was about 9.49 seconds.

Answer 14

I don't see 3C question

Answer 15

Let me get it really quick, I forgot to screenshot it.

Answer 16

I was guessing that was what it was asking. But your heights for t=3 and bigger are wrong. Remember, you should see the height be 0 about 9 seconds in
and your numbers are shooting up.
The good news is t=0, 1, and 2 have the correct heights.

Answer 17

So, what would the other heights be? I think I did it wrong. :/

Answer 18

I mean 24.

Answer 19

-16*3^2 + 150*3 + 18

Answer 20

24.

Answer 21

Oh wait. My bad.

Answer 22

324?

Answer 23

yes

Answer 24

And 4 would be 362?

Answer 25

yes

Answer 26

So, now that I fixed the table for heights, what am I supposed to do next?

Answer 27

what did you get for t=9
?

Answer 28

I got, 72.

Answer 29

ok, and at t=10 you will get -82 (I guess it fell into a lake and kept going down)
so that makes sense, somewhere between 9 and 10 seconds it was a ground level.

Answer 30

I don't understand why it is asking for the axis of symmetry when I don't have a graph.

Answer 31

3d. Axis of symmetry
this is the t where the parabola peaks.
it happens at t= -b/(2a)
where a,b,c are the numbers in front of t^2, t , and c is 18

Answer 32

I'm confused as to which numbers are going in the equation.

Answer 33

Something like: t=-16/(2a)?

Answer 34

the "convention" when we write a parabola in standard form is
y= a x^2 + b x + c
or in your case, t instead of x
but the a, b, and c are the numbers
in your case -16, 150, and 18

Answer 35

So, -16/(2*150)?

Answer 36

the same letters we use when we write down the quadratic formula

Answer 37

b is the 2nd number. you have your a and b swapped

we want t= -b/(2a)

Answer 38

t=-150/(2*-16)?

Answer 39

yes

Answer 40

Then all I do is solve it?

Answer 41

simplify it.

Answer 42

I got 4.6875

Answer 43

yes, so the line t= 4.7 (rounded) is the axis of symmetry

Answer 44

Ohh, okay, that makes sense. What about the next part?

Answer 45

the coordinates of the vertex (t,h)
where t is what you just found, and h you find (just like in your table) but use t=4.6875 (always use lots of decimals when calculating, then round the height at the end)

Answer 46

So, am I doing all of the numbers in the table?

Answer 47

For the table, you should figure out the heights for t=0 up to 10
(at 10 the rocket has gotten back to the earth)

For 3e
to find the coordinates of the vertex, you do the same thing you did for the table
but use t= 4.6875

Answer 48

Is there an equation I should use?

Answer 49

yes, the one in 3a

Answer 50

So, I'm just plugging in the numbers to this equation?
H(T) = -16T^2 + 150T + 18

Answer 51

yes, you are putting in t=4.6875
to find the height at that time. That will be the high point (vertex) of the parabola

Answer 52

H(T) = -16T^2 + 150T + 18
H(T) = -16(4.6875^2) + 150 * 4.6875 + 18?

Answer 53

yes

Answer 54

I got 369.5625.

Answer 55

although in theory you would write
H(4.6875)
which means H at t=4.6875

Answer 56

yes, that looks good,
so the vertex is at
(4.69, 369.56)

Answer 57

Is that what I would write?

Answer 58

yes

Answer 59

What about the next part? I'm also stuck there as well.

Answer 60

Part 3F?

Answer 61

if you shoot off the rocket at t=0
what would negative t be ?

Answer 62

I do not believe so.

Answer 63

pretend we have a stop watch and start it when we shoot the rocket
when it reads t=4.69 seconds we know the rocket is at its peak
and at t= 9.49 seconds it hits the ground

where is the rocket at t= -1 ?

Answer 64

t=-1 means 1 second before we shoot off the rocket

Answer 65

-116?

Answer 66

maybe according to the equation
but in the real world, the rocket is on the building waiting to go.

Answer 67

Oh, so that is why negative wouldn't make sense for the equation?

Answer 68

yes, the idea is the equation works only for time from t=0 until t=9.49
assuming the rocket hits the earth (no lake), it can't go to below the surface like your table shows.

Answer 69

There is another few parts I need help with, but I need to take a screenshot.

Answer 70

what happened to 4
?

Answer 71

One second, I forgot to screenshot that, my mistake.

Answer 72

For 4. you need a new equation that starts on the ground instead of on top of the building

Answer 73

I got number 5 and 6, thank you for your help!(:

Answer 74

@phi I need your help! I got confused and messed up, can you help me with 5 and 6?

Answer 75

I also need help with 4 too, I thought I had it, but I didn't.

Answer 76

For 4. you need a new equation that starts on the ground instead of on top of the building

Answer 77

How would I make that equation?

Answer 78

I mean I did make an equation, do you want to see it?

Answer 79

they say use the same initial velocity as you used in Task 1
what velocity did you use in Task 1?

Answer 80

I never used a velocity because I didn't know what it was. I was really confused..

Answer 81

ok, so in problem 4 you are using 100 ft/sec
hopefully that is ok.
your equation looks good, except people don't write +0
just write
H(t)= -16t^2+100t

Answer 82

Okay, so what would I do after that?

Answer 83

I would first find the time it reaches the peak (at the vertex)
remember how to do that ? t= -b/(2a)

Answer 84

When I did the equation, I found that the maximum height was about 3 or less.

Answer 85

we want to make sure it is bigger than 3
then we know at t=3 the rocket is still going up

Answer 86

Can I show you the rest of my equation? So you can see what I did?