Question

You launch a water balloon. The function h=-0.08t^2+1.6t+2 models the height h (in feet) of the water balloon after t seconds. After how many seconds is the water balloon at 9 feet?

Answer 1

6.5 seconds and 13.5 seconds

Need to know:

Quadratic formula:
`\frac{-b +-\sqrt{b^(2) - 4ac } }{2a}`

Explanation:

We have -0.08t² + 1.6t + 2 = 9

Subtract 9 from both sides

-0.08t² + 1.6t + 2 = 9

- 9 - 9

-0.08t² + 1.6t - 7 = 0

We will have to use the quadratic formula

a = -0.08

b = 1.6

c = -7

`\frac{-1.6 +-\sqrt{1.6^(2) - 4(-0.08)(-7) } }{2(-0.08)}`

`(-1.6 +-√(2.56 - 2.24 ) )/(-0.16)`

`(-1.6 +-√(0.32 ) )/(-0.16)`

`(-1.6 +√(0.32 ) )/(-0.16) = 6.464466`

`(-1.6 -√(0.32 ) )/(-0.16)=13.5355`

These are the two times the balloon will be at 9 feet.