Answer:
The cannonball lands at approximately 5.093 unit distance from the point of fire
Step-by-step explanation:
The given parameters are;
The arc denoting the equation of motion of the cannon is y₁ = -0.5·x² + 2.5·x + 1
The slope of the field where in the direction the cannon is fired is y₂ = 1.5·x
The points where the cannonball land on the slopping field is given as rightly pointed by equating the two equations, the cannonball path path and the field path as follows;
At the point of contact of the cannonball and the field, the y-values of both equation will be equal
y₁ = y₂
∴ -0.5·x² + 2.5·x + 1 = 0.15·x
Which gives;
-0.5·x² + 2.5·x - 0.15·x + 1 = 0
-0.5·x² + 2.35·x + 1 = 0
-(-0.5·x² + 2.35·x + 1) = 0.5·x² - 2.35·x - 1 = 0
0.5·x² - 2.35·x - 1 = 0
The above equation is in the general form of a quadratic equation, which is given as follows;
a·x² + b·x + c = 0
By the quadratic equation, we have;
[tex]x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}[/tex]
Plugging in the values, gives;
[tex]x = \dfrac{2.35\pm \sqrt{(2.35)^{2}-4\cdot (0.5)\times (-1)}}{2\cdot (0.5)} = \dfrac{2.35\pm \sqrt{7.5225}}{1} =2.35 \pm \sqrt{7.5225}[/tex]
∴ x ≈ 5.093 or x ≈ -0.393
Therefore, the cannonball will takeoff at x ≈ -0.393 and land at x ≈ 5.093
The height from which they fire the cannon is given by the substituting the value of x ≈ -0.393 into the equation for the path of the cannonball, to give;
[tex]y_{(initial)}[/tex] = -0.5·(-0.393)² + 2.5·(-0.393) + 1 = -0.0597
[tex]y_{(initial)}[/tex] ≈ -0.0597.
However, the actual initial height from which the cannonball is fired given by placing x = 0, which gives y = 1, which is the reason for the other (negative) value for x. Please see the attached graph created with Microsoft Excel.
