6. 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=-0.04x2 + 4x + 1, where y represents the height of the ball in meters and x represents the horizontal distance of the ball from the cannon. The incline of the hill can be represented by the equation y = 0.8125x. How far will the cannon ball have moved horizontally from the cannon when it hits the hill?

Its canon is placed on the floor, as we knew from the very first part. So, for Canon, [tex]y = 0[/tex].
Consider the origin, which is located at the bottom of a hill. So, to find the canon's x value, we must solve the quadratic formula: [tex]-0.04x^2 + 4x + 1 = 0[/tex] ,
which has only one negative solution,[tex]0.25.[/tex] We know that the x value cannot be positive as this would suggest that the cannon is located inside of the hill.
This hill's equation is[tex]y = 0.8125x[/tex] , and we understand that now the cannonball will collide with it at the junction of the hill line and its hyperbolic path.
So, when the formulas are combined, we get

[tex]\to 0.8125x =-0.04x2 + 4x + 1\\\\\to 0.04x^2 - 3.1875 - 1 = 0\\\\[/tex]

It has the only positive root as [tex]80[/tex]. As a result, the y value is [tex]0.8125\times 80 = 65[/tex]
The horizontal distance will be the x value + the [tex]0.25 = 80.25[/tex]
Its horizontal distance = x value+[tex]0.25= 80+0.25 = 80.25[/tex]

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