Question

A military 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 = -0.05x² + 5x + 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 = 0.725x. How far will the cannon ball have
moved horizontal from the cannon when it hits the hill?

Answer 1

9514 1404 393

Answer:

85.7 meters

Explanation:

A graphing calculator finds the result easily. The ball moved 85.7 meters horizontally before it hit the hill.

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Equating y expressions gives ...

0.725x = -0.05x^2 +5x +1

In standard form, this is ...

0.05x^2 -4.275x -1 = 0

The quadratic formula tells us the positive solution is ...

x = (4.275 +√(4.275^2 -4(0.05)(-1)))/(2(0.05)) = (4.275 +√18.475625)/.1

x ≈ 85.733

The cannon ball hits the hill at a horizontal distance of 85.7 meters from the cannon.