a cannon ball is launched into the air with a...
a cannon ball is launched into the air with an upward velocity of 383 feet per second, from a 6 - foot tall cannon. the height h of the cannon ball after t seconds can be found using the equation h = -16t² + 383t + 6. approximately how long will it take for the cannon ball to be 2218 feet high? round answers to the nearest tenth if necessary. units:
Answer
# Explanation:
## Step1: Set up the equation
Set $h = 2218$ in the equation $h=-16t^{2}+383t + 6$. So we get $-16t^{2}+383t + 6=2218$. Rearrange it to the standard - quadratic - form $ax^{2}+bx + c = 0$. We have $-16t^{2}+383t+6 - 2218 = 0$, which simplifies to $-16t^{2}+383t - 2212 = 0$. Multiply through by - 1 to get $16t^{2}-383t + 2212 = 0$.
## Step2: Use the quadratic formula
The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 16$, $b=-383$, and $c = 2212$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-383)^{2}-4\times16\times2212=146689-141568 = 5121$.
## Step3: Calculate the values of t
$t=\frac{383\pm\sqrt{5121}}{32}$. $\sqrt{5121}\approx71.56$. Then $t_1=\frac{383 + 71.56}{32}=\frac{454.56}{32}\approx14.2$ and $t_2=\frac{383 - 71.56}{32}=\frac{311.44}{32}\approx9.7$.
# Answer:
$t\approx9.7$ seconds or $t\approx14.2$ seconds