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

Answer:

6.5 seconds and 13.5 seconds

Need to know:

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

Step-by-step 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

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

$\frac{-1.6 +-\sqrt{2.56 - 2.24 } }{-0.16}$

$\frac{-1.6 +-\sqrt{0.32 } }{-0.16}$

$\frac{-1.6 +\sqrt{0.32 } }{-0.16} = 6.464466$

$\frac{-1.6 -\sqrt{0.32 } }{-0.16}=13.5355$

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