Question

Algebra 1 question :D
Medal + Fan!

Answer 1

The function H(t) = -16t^2 + 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.

Question: What is the maximum height that the projectile will reach? Show your work.

Answer 2

@AlexandervonHumboldt2 @bibby

Answer 3

@paki

Answer 4

you have no numbers to work with here so your answer must be in terms of \(v\) and \(s\)

Answer 5

ok will do...

Answer 6

which is kind of annoying, but we can do it

Answer 7

what shall I begin with?

Answer 8

first coordinate of the vertex is always \(-\frac{b}{2a}\) which in your case is \[\frac{v}{32}\]

Answer 9

ok...

Answer 10

second coordinate (the max you are looking for) is what you get when you replace \(t\) by \(\frac{v]{32}\)

Answer 11

oops
replace \(t\) by \(\frac{v}{32}\)

Answer 12

nah thats ok..... hmm, now how would you replace t by (v/32)?

Answer 13

(t/32)?

Answer 14

no just plug it in

Answer 15

no wait.... that would just literally mean replace

Answer 16

yeah, yeah... i'm a wee bit slow

Answer 17

http://www.wolframalpha.com/input/?i=-16%28v%2F32%29^2%2Bv*v%2F32%2Bs

Answer 18

s+((v^2)/(64))

Answer 19

this is point 2, not max height right?

Answer 20

that is the max height yes

Answer 21

depends on both s and v

Answer 22

thanks how would I show my work?

Answer 23

Since first coordinate of the vertex is always −((b)/(2a)) which in my case is
((v)(32)). To find the max, you have to replace t by ((v)(32)). Then you just plug t in, to get s+((v^2)/(64)). Thus, the max height is s+((v^2)/(64))

Answer 24

looks good to me

Answer 25

Thanks... can you live through 2 more questions?

Answer 26

lol maybe lets see

Answer 27

Both of these have to do with the original statement.

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]

Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?

Answer 28

\[g(t)=31+32.2t\]?

Answer 29

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.

That's all I got!

Answer 30

ok then i think we are missing something
the first line about H has this
The function H(t) = -16t^2 + 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.

Answer 31

it has no numbers in it at all

Answer 32

so without knowing \(v\) or \(s\) we can't possibly do this problem
we can't see where they intersect
i am thinking somewhere along the line you were told v and s, and we missed it somehow

Answer 33

if we do know v and s we have to change the answer to the first question too

Answer 34

Yeah.... but remember this is part c and d of the question we just answered.

Answer 35

The projectile was launched from a height of 96 feet with an initial velocity of 80 feet per second.

Answer 36

FOUND IT :D

Answer 37

oh lord have mercy
ok lets start at the top (quickly)

Answer 38

ok then

Answer 39

haha... "oh lord have mercy"

Answer 40

forget the first answer with the s and v, now we do it with numbers
\[h(t)=-16t^2+80t+96\]
first coordianate is \[-\frac{b}{2a}\]which is now
\[\frac{80}{32}=2.5\]

Answer 41

second coordinate, which is the maximum, is \(h(2.5)\)which i will let you compute (use a calculator )

Answer 42

so 2.5 = −16t^2+80t+96

or

16*2.5^2+80*2.5+96

Answer 43

\[g(t)=31+32.2t\] they will intersect when \(h(t)=g(t)\) so we set them equal and solve

Answer 44

you are missing a minus sign \[-16*(2.5)^2+80*2.5+96
\]

Answer 45

i get 196
http://www.wolframalpha.com/input/?i=-16*2.5^2%2B80*2.5%2B96

Answer 46

ok then so for the first one we did.... the answer is 196?

Answer 47

yes

Answer 48

max height is 196 when \(t=2.5\)

Answer 49

Mind helping me put this answer in to a few sentences?

Answer 50

say exactly what you said before, but use the numbers rather than the s and t

Answer 51

oh ok... gimme a sec then :P

Answer 52

Since first coordinate of the vertex is always −((b)/(2a)) which in my case is 80/32=2.5. To find the max, you have to replace t by 2.5. Then you just plug t in, to get -16(2.5)^2+80*2.5=96=196 Thus, the max height is 196 feet

Answer 53

ok now for C

Answer 54

Ignore my reply, that was for the first answer :P

Answer 55

set
\[h(t)=-16t^2+80t+96=g(t)=31+32.2t\] and solve for \(t\)

Answer 56

it says use a table, we will use technology

Answer 57

haha... ok then :D

Answer 58

http://www.wolframalpha.com/input/?i=-16t^2%2B80t%2B96%3D31%2B32.2t+domain+0..5

Answer 59

we graph them both we see that they do intersect, somewhere around \(t=4\) i.e. in 4 seconds

Answer 60

How would you find the approximate solution to the equation H(t) = g(t)?

−16t^2+80t+96=31+32.2t

Answer 61

set them equal and ask wolfram

Answer 62

ok

Answer 63

Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?
since they meet around 4 second, the projectile is going down
that is because it reaches it's maximum height at 2.5 seconds and starts going down after

Answer 64

t=−1.0149925261473127,4.0024925261473125 thats what I got...

Answer 65

i got 4

Answer 66

did you check the link i sent ?

Answer 67

yeah, but I entered it incorrectly :P

Answer 68

so they intersect at 2?

Answer 69

x = 2?