the path of a football thrown across a field is given by the equation y=-0.005x^2 + x+ 5, where x represents the distance, in feet, the ball has traveled horizontally and y represents the height, in feet, of the ball above level ground
A) About how far has the ball traveled horizontally when it returns to the ground?
B) What is the maximum height the ball reaches?
A) About how far has the ball traveled horizontally when it returns to the ground?
B) What is the maximum height the ball reaches?
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a) y=-0.005x^2 + x + 5
The ball lands at it's furthest point, when y = 0, implying 0 height.
.: When y=0, => 0 = -0.005x^2 + x +5
Recognising the equation as a quadratic (Ie: ax^2 + bx + c),we can apply the Quadratic Formula to find a solution for x when y = 0.
Quadratic Formula: [-b +- sqrt(b^2 - 4*a*c)]/2*a
a = -0.005x^2, b = 1, c = 5
Substitute these values into their respective place within the Quadratic Formula and you get:
.: [-1 +- sqrt(1^2 - 4*(-0.005)*(5))]/2*(-0.005)
=[-1 +- sqrt(1+0.1)]/ -0.01
Using a calulator, we obtain answers of x = -4.88 and x = 204.881.
Assuming you cannot have negative (-) distance, we assume the answer to be X = 204.881.
.: The horizontal distance covered by the ball is 204.881m
b) The maximum height of a parabola, is found via the derrivative (dy/dx) and making it = 0.
If y = -0.005x^2 + x +5,
=> dy/dx = -0.01x + 1
When dy/dx = 0, => 0 = -0.01x +1
.: -1 = -0.01x
.: 1 = 0.01x
.: x = 1/0.01
= 100
However that is only the X value. To obtain the greatest height, you must substitute that value back into the original equation to obtain the height.
.: when x = 100, y = -0.005*(100)^2 + 100 +5
= 55m
The ball lands at it's furthest point, when y = 0, implying 0 height.
.: When y=0, => 0 = -0.005x^2 + x +5
Recognising the equation as a quadratic (Ie: ax^2 + bx + c),we can apply the Quadratic Formula to find a solution for x when y = 0.
Quadratic Formula: [-b +- sqrt(b^2 - 4*a*c)]/2*a
a = -0.005x^2, b = 1, c = 5
Substitute these values into their respective place within the Quadratic Formula and you get:
.: [-1 +- sqrt(1^2 - 4*(-0.005)*(5))]/2*(-0.005)
=[-1 +- sqrt(1+0.1)]/ -0.01
Using a calulator, we obtain answers of x = -4.88 and x = 204.881.
Assuming you cannot have negative (-) distance, we assume the answer to be X = 204.881.
.: The horizontal distance covered by the ball is 204.881m
b) The maximum height of a parabola, is found via the derrivative (dy/dx) and making it = 0.
If y = -0.005x^2 + x +5,
=> dy/dx = -0.01x + 1
When dy/dx = 0, => 0 = -0.01x +1
.: -1 = -0.01x
.: 1 = 0.01x
.: x = 1/0.01
= 100
However that is only the X value. To obtain the greatest height, you must substitute that value back into the original equation to obtain the height.
.: when x = 100, y = -0.005*(100)^2 + 100 +5
= 55m