The height h (in feet) of a volleyball t seconds after it is hit can be modeled by h = -16t^2+48t+4. How long is the volleyball in the air?

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Hugo C. Share
Nov 15, 2015

$t = 3.08 s$

Explanation:

We have to substitute $h$ by $0$, as that is the height both when projected as when it hits the ground. Then we solve for $t$:
$- 16 {t}^{2} + 48 t + 4 = 0$,
$16 {t}^{2} - 48 t - 4 = 0$,
$4 \left(4 {t}^{2} - 12 t - 1\right) = 0$,
$4 {t}^{2} - 12 t - 1 = 0$. Now we use $t = \frac{- b \pm \sqrt{\left({b}^{2} - 4 a c\right)}}{2 a}$:
$t = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(4\right) \left(- 1\right)}}{2 \left(4\right)}$,
$t = \frac{12 \pm \sqrt{144 + 16}}{8}$,
$t = \frac{12 \pm \sqrt{160}}{8}$,
$t = 3.08 s$.

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