Question

a projectile is thrown into the air. the equation y=-3x^2-12x+96 represents the path of the projectile, where x represents the amount of time in seconds. how long is the projectile in the air, in seconds

Answer 1

To determine how long the projectile is in the air, we need to find when it reaches the ground, i.e., when \(y = 0\) in the equation \(y = -3x^2 - 12x + 96\).

We set up the equation:

\[
-3x^2 - 12x + 96 = 0
\]

To simplify, we can divide each term by -3:

\[
x^2 + 4x - 32 = 0
\]

Next, we will use the quadratic formula to find the values of \(x\). The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our case, \(a = 1\), \(b = 4\), and \(c = -32\). We'll first calculate the discriminant:

\[
b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-32) = 16 + 128 = 144
\]

Now we can find \(x\):

\[
x = \frac{-4 \pm \sqrt{144}}{2 \cdot 1}
\]
\[
x = \frac{-4 \pm 12}{2}
\]

This gives us two solutions:

1.
\[
x = \frac{-4 + 12}{2} = \frac{8}{2} = 4
\]

2.
\[
x = \frac{-4 - 12}{2} = \frac{-16}{2} = -8
\]

Since time cannot be negative, we discard \(x = -8\).

Thus, the projectile is in the air for \( \boxed{4} \) seconds.