Question

A firecracker shoots up from a hill 160 feet high, with an initial speed of 90 feet per second. Using the formula H(t) = −16t^2 + vt + s, approximately how long will it take the firecracker to hit the ground?

Five seconds
Six seconds
Seven seconds
Eight seconds

Answer 1

We need to substitute values that are given to us by the context

v is velocity, what is the speed given to us here?

Answer 2

90 ft per second

Answer 3

\(H(t) = −16t^2 + vt + s\) we substitute v for 90

\(H(t) = −16t^2 + 90t + s\)

The s is the original height, what was the original height given to us here?

Answer 4

Original height would be 160 ft high

Answer 5

Yes
So our equation is:

\(H(t) = −16t^2 + 90t + 160\)

When it says when the firecracker reaches the ground, we want the height to be 0 (because its in the ground)

We set this equal to 0
\(0 = −16t^2 + 90t + 160\)

Answer 6

okay got it down

Answer 7

Do you know how to solve for t?

Answer 8

I have no idea what t would end up being

Answer 9

Okay we can use the quadratic equation, do you know what that is?

Answer 10

I am supposed to but i honestly forgot and i cant find my notebook

Answer 11

I am using a spare notebook until i can find the one from yesterday

Answer 12

\(t=\large \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Have you seen this before?

Answer 13

o.o ive seen it but never figured out how to use it

Answer 14

Okay so we need to find the a, b and c values for them

A quadratic equation is written as \(y=ax^2+bx+c\)

With what we have \(H(t)=-16t^2+90t+160\)

What are the a, b and c

Answer 15

A: -16
B: 90
C:160
Right.....?

Answer 16

Yes

Answer 17

Now we substitute what we have into this
\(a=-16\)
\(b=90\)
\(c=160\)

\(t=\large \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 18

Alright

Answer 19

So now we just substitute them?

Answer 20

Yes

Answer 21

Okay one sec

Answer 22

Idk how to do the thing that you did with the equation so i can show you.....

Answer 23

`\(t=\large \frac{- b \pm\sqrt{b ^2-4 a c}}{2 a}\)`
Copy that and insert the a,b and c

Answer 24

\(t=\large \frac{- -90 \pm\sqrt{90 ^2-4 (-16) (160)}}{2 -16}\)

Answer 25

Yes, double negatives will make a positive
--90 = 90

And multiply what is in within the root

Multiply the 2 and -16

Answer 26

-32

Answer 27

Right so we have
\(t=\large \frac{90 \pm\sqrt{90 ^2-4 (-16) (160)}}{-32}\)

Multiply what is inside the square root now

\(t=\large \frac{90 \pm\sqrt{\color{red}{90 ^2-4 (-16) (160)}}}{-32}\)

Answer 28

okay one sec

Answer 29

so like 90^2 = 8100? and would i do -4 * -16 and -4 * 160?

Answer 30

Yes \(90^2\) is 8100 and then you multiply -4 times -16 and then the answer between those times 160

Answer 31

So 64-160?

Answer 32

\(64 \times 160\)

Answer 33

Ohh okay one sec

Answer 34

so that would equal 10240

Answer 35

Right
Now add 8100 and 10240

Answer 36

18340

Answer 37

Right so we have
\(\large \frac{90\pm\sqrt{18340}}{-32}\)

Find \(\sqrt{18340}\) using your calculator

Answer 38

thats square root of 18340?

Answer 39

Yes

Answer 40

135.425256138

Answer 41

You can just round it do \(135.425\)

Basically now your answer can be:

\(\large \frac{90+135.425}{-32}\)

or
\(\large \frac{90-135.425}{-32}\)

Solve both

Answer 42

225.425/-32 or -45.425/-32

Answer 43

Yes, divide them

Answer 44

okay one sec

Answer 45

-7.044 or 1.419

Answer 46

Ah made a sign mistake

\(\large \frac{-90+135.425}{-32}\)

or
\(\large \frac{-90-135.425}{-32}\)

Those answers are right, the signs are just flipped

\(7.04~and~-1.419\)

Answer 47

Okay so from here the question asks for time

We cannot have negative time so \(-1.419\) is deleted


\(7.04\) is close to 7 seconds so 7 is your answer

Answer 48

Ohh okay thank you. You really have been helping me alot.

Answer 49

Of course, thats what I am here for

Answer 50

^~^

Answer 51

:)