Question

a civil war cannon is placed at the base of a hill. the cannon is fired at an angle toward the hill. the path of the cannon ball is parabolic and can be represented by ( y = - 0.04 x ^ { 2 } + 4 x + 1 ), where y represents the height of the ball in feet and x represents the horizontal distance of the ball from the cannon. the incline of the hill can be represented by the equation ( y = 0.8125 x ). how far will the cannon ball have moved horizontally from the cannon when it hits the hill? record your answer. be sure to use the correct place value.

Explanation:

Step1: Set the two equations equal

Since the cannon - ball hits the hill when \(y\) (height) is the same for both the cannon - ball's path (\(y=-0.04x^{2}+4x + 1\)) and the hill's incline (\(y = 0.8125x\)), we set \(-0.04x^{2}+4x + 1=0.8125x\).

Step2: Rearrange the equation to standard quadratic form

Subtract \(0.8125x\) from both sides: \(-0.04x^{2}+4x-0.8125x + 1=0\).
Simplify to get \(-0.04x^{2}+3.1875x + 1=0\). Multiply through by \(-1\) to make the coefficient of \(x^{2}\) positive: \(0.04x^{2}-3.1875x - 1=0\).
For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 0.04\), \(b=-3.1875\), \(c=-1\)), we can use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step3: Calculate the discriminant

First, calculate \(b^{2}-4ac\):
\(b^{2}-4ac=(-3.1875)^{2}-4\times0.04\times(-1)\)
\(=10.16015625 + 0.16\)
\(=10.32015625\)

Step4: Apply the quadratic formula

\(x=\frac{-(-3.1875)\pm\sqrt{10.32015625}}{2\times0.04}=\frac{3.1875\pm3.2125}{0.08}\)
We have two solutions for \(x\):
\(x_1=\frac{3.1875 + 3.2125}{0.08}=\frac{6.4}{0.08}=80\)
\(x_2=\frac{3.1875-3.2125}{0.08}=\frac{-0.025}{0.08}=-0.3125\)

Since \(x\) represents a horizontal distance (and distance cannot be negative in this context), we discard \(x=-0.3125\).

Answer:

\(80\)