Question

a balloon ascending at a rte of 12ft/s is at a height of 80ft above the ground when a package is dropped. How long does it take the package to reach the ground?

Answer 1

\[a(t) = -32\]
\[v(t) = -32t + c\]

Answer 2

the c is 12
so \[v(t) = -32t + 12\]
\[s(t) = -16t^2 + 12t + d\]
and d = 80
so
\[s(t) = -16t^2 + 12t + 80\]

Answer 3

This doesn't look like its going to reduce, nicely, I think I did something wrong. Also I forgot to mention that acceleration due to gravity is 32ft/s

Answer 4

@Hero

Answer 5

@TheSmartOne

Answer 6

@robtobey2

Answer 7

@Jadeishere

Answer 8

I'm not the best person to ask! Terribly sorry!
@mathmate

Answer 9

I think I just need to reduce that last part and solve for 0 but I am not sure how.

Answer 10

quadratic formula

Answer 11

\[ax^2 + bx + c = 0\]
\[x = \frac{ -b +-\sqrt{b^2 - 4ac} }{ 2a }\]

Answer 12

Just use the equation
S=ut+ 1/2 a t^2
U = 12ft/s
S = displacement = 80 ft
a = -32ft/s^2
Find t

Answer 13

80 = 12t - 16t²
16t²-12t+80 = 0
Solve this

Answer 14

right, that is the issue...

Answer 15

okay, but I don't know this part:
\[16^{2} + 12t = -80\]

Answer 16

I think I could just use the quadratic formula

Answer 17

\[t = \frac{ -12 +- \sqrt{12^2 - 4 (-16)(80)} }{ 2(-16) }\]

Answer 18

Are you sure the numbers are correct/?

Answer 19

are you asking because you know the numbers are incorrect?

Answer 20

Well because the answer is coming in imaginary form

Answer 21

yeah, I am pretty sure the numbers are correct, but the 16t²-12t+80 = 0 can reduce to 4t^2 - 3t + 20 = 0

Answer 22

solved! \[-3+ \sqrt{\frac{ 329 }{ 8 }}\]

verified with solution.

Answer 23

also with wolframalpha: http://www.wolframalpha.com/input/?i=0+%3D+-16x%5E2+%2B+12x+%2B+80,+find+x

Answer 24

Oh yep got it

Answer 25

because \[t = -3 + \sqrt{\frac{ -3^2 - 4(4)(20) }{ 2(4) }}\]

goes to \[t = -3 + \sqrt{\frac{ -9 - 320}{ 8 }}\]

Answer 26

If we use upwards as positive, the distance travelled is -80 feet, and the initial speed is +12 ft/s.
This comes out to be
\(16t^2-12t-80=0\)
and the quadratic formula will give two real solutions. Reject the negative value.
The solution retained should be between 2 and 3 seconds.