Question

1Nonlinear Systems: To prevent large avalanches of snow in mountainous areas, explosive shells
can be launched to trigger carefully placed smaller avalanches, A shell is launched on a path
modeled by a quadratic equation y = - 0. 16 x2 + x, where the distances are in kilometers. The
mountainside can be modeled by a linear function y = 0.75 x - 2. Where does the shell hit?
A) The shell hits about 1.3 km high on the mountain side at a horizontal distance away of about
4.4 km.
B) The shell hits about 4.4 km high on the mountain side at a horizontal distance away of about
1.3 km.
C) The shell hits about 2.1 km high on the mountain side at a horizontal distance away of about
3.4 km.
D) The shell hits about 3.4 km high on the mountain side at a horizontal distance away of about
2.1 km.

Answer 1

To find where the shell hits the mountainside, we need to set the two equations equal to each other. The equations are:

1. Shell's trajectory: \( y = -0.16x^2 + x \)
2. Mountainside: \( y = 0.75x - 2 \)

Set the two expressions for \( y \) equal to each other:

\[
-0.16x^2 + x = 0.75x - 2
\]

Now, rearranging this equation gives:

\[
-0.16x^2 + x - 0.75x + 2 = 0
\]

Simplifying:

\[
-0.16x^2 + 0.25x + 2 = 0
\]

To make calculations easier, multiply the entire equation by -1 to eliminate the negative coefficient:

\[
0.16x^2 - 0.25x - 2 = 0
\]

Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 0.16 \), \( b = -0.25 \), and \( c = -2 \).

Calculate the discriminant:

\[
b^2 - 4ac = (-0.25)^2 - 4(0.16)(-2) = 0.0625 + 1.28 = 1.3425
\]

Now, calculate the values of \( x \).

\[
x = \frac{-(-0.25) \pm \sqrt{1.3425}}{2(0.16)} = \frac{0.25 \pm \sqrt{1.3425}}{0.32}
\]

We first find \( \sqrt{1.3425} \):

\[
\sqrt{1.3425} \approx 1.16
\]

Now substituting this value back in, we can calculate the two possible values for \( x \):

\[
x_1 = \frac{0.25 + 1.16}{0.32} \approx \frac{1.41}{0.32} \approx 4.41 \text{ km}
\]
\[
x_2 = \frac{0.25 - 1.16}{0.32} \approx \frac{-0.91}{0.32} \approx -2.84 \text{ km} \, (\text{not applicable since distance cannot be negative})
\]

So we take the \( x \) value of approximately \( 4.41 \) km.

Now to find the height \( y \) when \( x = 4.41 \) km, substitute back into either equation.

Using the mountainside equation:

\[
y = 0.75(4.41) - 2
\]
\[
y = 3.3075 - 2 = 1.3075 \approx 1.3 \text{ km}
\]

Thus, the shell hits the mountainside at about **1.3 km high** and at a horizontal distance of about **4.4 km** away.

So the correct answer is:

**A) The shell hits about 1.3 km high on the mountainside at a horizontal distance away of about 4.4 km.**