Question

Multiple hard Algebra 2 questions
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The pilot of a helicopter plans to release a bucket of water on a forest fire. The height y in feet of the water t seconds after its release is modeled by y = -16t^2 - 2t + 400. The horizontal distance x in feet between the water and its point of release is modeled by x = 91t. To the nearest foot, at what horizontal distance from the target should the pilot begin releasing the water? in feet
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If a tightrope walker falls, he will land on a safety net. His height h in feet after a fall can be modeled by h(t) = 60 - 16t2, where t is the time in seconds. If the safety net is 11 feet above the ground, how many seconds will the tightrope walker fall before landing on the safety net? Round your answer to the nearest hundredth. in seconds

Answer 1

do they tell you the height of the helicopter ?

Answer 2

no, this is all they give me :-/

Answer 3

I guess they want you to figure out "t" when y=0
y=0 means the water is at height 0. you want to know how long it takes.
once you find t, use it in x = 91*t to find x

so step 1 is solve for t in
-16t^2 - 2t + 400=0

Answer 4

so do I use the quadratic equasion then? @phi

Answer 5

that looks like your best approach.
I would first divide the equation (all terms, both sides) by 2, to make it just a little simpler.

you get
-8 t^2 -t + 200 = 0/2 or
\[ -8 t^2 -t + 200 = 0 \]

a= -8, b= -1, c =+200

Answer 6

\[x=1\pm \frac{ \sqrt{-1^{2}-4\times-8\times200}\ }{ 2\times-8 }\]

Answer 7

ok, except the 2*-8 divides the 1 also.

Answer 8

\[x=1\pm \frac{-6399 }{ -16 }\]

Answer 9

what do uou mean "ok, except the 2*-8 divides the 1 also."?

Answer 10

\[x =-b \pm \frac{ \sqrt{b ^{2-4\times a \times c}} }{ 2a }\]

Answer 11

\[ x= \frac{ 1\pm\sqrt{(-1)^{2}-4\times-8\times200}\ }{ 2\times-8 } \]

Answer 12

see http://www.purplemath.com/modules/solvquad4.htm

Answer 13

now inside the square root, you should get
-1*-1 + 4*8*200

Answer 14

oh, ok so from there it is \[x =1\pm \frac{ \sqrt{1+6,400} }{ -16 }\]

Answer 15

\[=1\pm \frac{ \sqrt{6,401} }{ -16 }\]

Answer 16

you are a bit stubborn, aren't you ??
it is
\[ = \frac{ 1\pm\sqrt{6,401} }{ -16 } \]

Answer 17

that cant be the final form, where does it go from there? where do I even begin to simplify it from there?

Answer 18

This one needs a calculator.

Answer 19

ok

Answer 20

only 1 of the solutions (the positive one) makes sense.

Answer 21

ok so \[\sqrt{6401} =80.0062497559\] so it would now read \[=\frac{ 1\pm80.0062497559 }{ -16 }\]

Answer 22

notice, to get a positive time, you want the top to be negative.
in other words choose
(1-80.0062497559)/-16

Answer 23

oh! ok, so the solutions are -5.063125 and 4.938125 so the 4.9 would be the best answer?

Answer 24

I would not round the time. only round after you get the final number.

Answer 25

oh ok so just input the whole 4.938125 then?

Answer 26

now find x

x = 91t

Answer 27

x = 449.369375

Answer 28

now do I round it? to 449.4

Answer 29

To the nearest foot, at what horizontal distance

Answer 30

oh ok so just to 449 then, ok

Answer 31

btw, you got t= 4.938125

however, the correct t is 4.9378906

Your answer was close enough, but I would re-do that step... just to make sure you can get the correct answer.

Answer 32

ok, my lesson said it was right the 449, ya, my iphone only calculates to the 7th place sadly, I need to get a real calculator eventually

Answer 33

it rounds what is after it

Answer 34

oh, not sure what it did there, thats the same amount of digits, I'll have to watch that with this thing

Answer 35

7 places is fine.... I just wanted to make sure you weren't doing something wrong.

As for the second problem, half of the "fun" is figuring out what equation to solve.

Can you try to write down the equation you should solve?

Answer 36

ya, I think so, it would be \[-16t ^{2}+60\] right?

Answer 37

that is not an equation. equations have an = sign in them

Answer 38

well, \[h(t)=-16t ^{2}+60\] then?

Answer 39

no, that equasion =0 then?

Answer 40

that tells you His height h in feet after a fall, after t seconds

what is his height when he hits the net ?

Answer 41

−16t^2+60=0

that would let you find the time it takes to hit the ground (the ground is at "height" zero)

Answer 42

First, we do not (yet) have the correct equation.
Second, you are *STILL* using the wrong formula for the quadratic equation.

Answer 43

what is his height when he hits the net ?

Answer 44

ya, the height of the net is 11 ft off the ground so does that mean that it is -11x in the equasion?

Answer 45

so the equasion reads -16t^2-11x+60=0?

Answer 46

This is important... h(t) = -16t^2 + 60

that tells you his height after t seconds.
for example, when t=0 (just when he starts to fall) his height is h(0)= -16*0+60
or 60 feet.

we know have some number of seconds his height will be less (he is falling)
in other words we know that h(t) = 11 feet
we want to find the time when that happens.

Answer 47

we know after some number of seconds his height will be less (he is falling)
in other words we know that at some point h(t) = 11 feet
we want to find the time when that happens.

Answer 48

ok so would 11 replace the 0 then? so -16t^2+60=11?

Answer 49

yes, exactly. I hope that makes sense.

Answer 50

ya, I think it does, so to make it quadratic able, I need to make it =0 so I take 11 from both sides?

Answer 51

-16t^2+49=0?

Answer 52

yes

Answer 53

ok so put that into quadratic equasion is \[x=\frac{ 0\pm \sqrt{b ^{2}-4\times-16\times49} }{ 2\times-16 }\]?

Answer 54

whoops should have been a 0 in that b, was too busy singin the song in my head!

Answer 55

that will work. fyi, notice when you don't have the "x" term, you can do this
\[-16t^2+49=0 \\ -16t^2 = -49 \\ t^2= \frac{49}{16} \]

Answer 56

which can be solved in your head (if the song is not confusing you)

Answer 57

t^2=2.0625

Answer 58

no, 3., hit the wrong key

Answer 59

3.0625

Answer 60

that is t^2

Answer 61

ya, so t is 1.75

Answer 62

of course sqr(49)/sqr(16) = 7/4
so you get a nice number for t= 1 ¾ or 1.75 exactly

Answer 63

t=1.75, so -16(1.75)^2+49=0, so 1.75 is my answer?

Answer 64

yes. they wanted it rounded to the nearest 100th, so 1.75 is the answer.

Answer 65

good job!

Answer 66

OK! GREAT! I think I get the concept now! If I have any questions on others I'll make sure to ask you, THANK YOU SOOOOO MUCH!!!!!!

Answer 67

An outfielder throws a baseball to the player on third base. The height h of the ball in feet is modeled by the function h(t) = -16t2 + 19t + 5, where t is time in seconds. The third baseman catches the ball when it is 4 feet above the ground. To the nearest tenth of a second, how long was the ball in the air before it was caught?

\[-16t^2+19t+5=4\]\[-16t^2+19t+1=0\]\[x=\frac{ -19\pm \sqrt{19^{2}-4\times-16\times1} }{ 2\times-16 }\]\[x=\frac{ -19\pm \sqrt{361-64} }{ -32 }\]\[x=\frac{ -19\pm \sqrt{297} }{ -32 }\]\[x=\frac{ -19\pm 17.2336879396}{ -32 }\]\[x=\frac{ -19 + 17.2336879396}{ -32 } = 0.05519725\]\[x=\frac{ -19 - 17.2336879396}{ -32 } = 1.13230275\]

@phi did I do it right here?

The answers dont seem to work out, they made the equasion = 2 and 2.09749553...

Answer 68

inside the square root you have 361 - 4*-16*1
notice that simplifies to 361 - (-64) = 361+64

Answer 69

so the inside is 425 instead then? oh! wow that threw my whole equasion off...

Answer 70

\[x=\frac{ -19+20.6155281281 }{ -32 } = 0.05048525\]
\[x=\frac{ -19-20.6155281281 }{ -32 } = 1.23798525\]

Answer 71

@phi

Answer 72

The first number should be negative (and obviously not the answer)

for the answer, round the second value the nearest tenth of a second

Answer 73

notice if we "plug in" t= 1.23798525 into the original equation
h(t) = -16t2 + 19t + 5
you will get h= 4 feet
which is the height at which the baseball was caught. So that looks good.

Answer 74

Oh! I was kinda confused because I thought it was supposed to make the equasion 0 but it is 4 ft above the ground! DUH whoops, thanks!