To find where the shell hits the mountainside, we need to set the two equations equal to each other. The equations are:
- Shell's trajectory: \( y = -0.16x^2 + x \)
- 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.