Question 749959
In the absence of air resistance, the trajectory of a cannon ball fired from ground level at 1200 feet per second and at an angle of 45 degrees can be approximated as
y=x-{{{1/45000}}}x^2
WRite this in the standard form: y = ax^2 + bx + c
y = {{{-1/45000}}}x^2 + x
where y is the height above the ground and x is the distance from the cannon.
A. Find the vertex of the parabola described by this equation
Find the axis of symmetry which is x = -b/(2a)
x = {{{(-1)/(2*(-1/45000))}}} = {{{(-1)/((-2/45000))}}} = +22500
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B. What is the maximum height of the cannon ball?
Replace x with 22500 in the original equation, find y
y = {{{-1/45000}}}22500^2 + 22500
y = +11250 ft max height
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C. What is the range of the cannon ball (the range is the horizontal distance to the point of impact at ground level)?
At impact, y = 0, occurs twice the value of the axis of symmetry
x = 2(22500) = 45000 ft
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Graphically
{{{ graph( 300, 200, -10000, 50000, -2000, 15000, (-1/45000)x^2+x) }}}