The path of a large stone fired from a torsion catapult can be modeled by y=-0.0054x(squared)+1.145x, where x is the distance the stone traveled (in yards) and y is the hieght of the stone (in yards)
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To determine the maximum height of the stone and the distance it will travel, we need to find the vertex of the quadratic equation y = -0.0054x² + 1.145x.
### Finding the Maximum Height ###
The equation is in the form of y = ax² + bx, where a = -0.0054 and b = 1.145. The vertex of a quadratic equation is given by the formula: x = -b / (2a).
Substituting the values, x = -1.145 / (2 * -0.0054), we can calculate the x-coordinate of the vertex.
x = -1.145 / (-0.0108) = 106.0185 (rounded to 4 decimal places)
Now, substitute this x value into the equation to find the corresponding y-coordinate:
y = -0.0054 * (106.0185)² + 1.145 * (106.0185)
y = 57.8822 (rounded to 4 decimal places)
Therefore, the maximum height of the stone is approximately 57.8822 yards.
### Finding the Distance Traveled ###
To find the distance the stone will travel, we can use the x-intercepts of the equation, which occur when y = 0.
0 = -0.0054x² + 1.145x
We can solve this quadratic equation for x using either factoring, completing the square, or the quadratic formula. However, in this case, our equation can be easily factored.
0 = x(-0.0054x + 1.145)
Setting each factor equal to zero, we get:
x = 0 (which represents the starting point, so we can ignore it in this context)
-0.0054x + 1.145 = 0
Simplifying this equation, we find:
-0.0054x = -1.145
x = -1.145 / -0.0054
x ≈ 212.0370 (rounded to 4 decimal places)
Hence, the stone will travel approximately 212.0370 yards.
In summary, the maximum height of the stone fired from the torsion catapult is approximately 57.8822 yards, and it will travel for approximately 212.0370 yards.