# A firework is launched from the ground at a velocity of 180 feet per second. Its height after t (t is the variable) seconds is given by the polynomial -16^2 + 180t. How do you find the height of the firework after 2 seconds and after 5 seconds?

Jun 13, 2015

In your equation I think there is a $t$ missing!!!
$h = 180 t - 16 {t}^{2}$

#### Explanation:

$h$ is the height reached starting with an initial velocity of $180 \frac{f t}{s}$ and considering an acceleration of gravity of $32 \frac{f t}{{s}^{2}}$:
$h \left(t\right) = {v}_{i} t - \frac{1}{2} \left(32\right) {\left(t\right)}^{2}$

$h \left(t\right) = 180 t - 16 {t}^{2}$

for:
$t = 2 s$
$h \left(2\right) = \textcolor{red}{296} f t$
$t = 5 s$
$h \left(5\right) = \textcolor{red}{500} f t$