SOLUTION: 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 45o can be approximated as
y=x-1/45,000x^2
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y=x-1/45,000x^2
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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 45o can be approximated as
y=x-1/45,000x^2
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
B. What is the maximum height of the cannon ball?
C. What is the range of the cannon ball (the range is the horizontal distance to the point of impact at ground level)?
I'm a bit confused on how to find all these! Any help is appreciated thanks! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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-x^2
WRite this in the standard form: y = ax^2 + bx + c
y = 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 = = = +22500
:
B. What is the maximum height of the cannon ball?
Replace x with 22500 in the original equation, find y
y = 22500^2 + 22500
y = +11250 ft max height
:
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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