SOLUTION: big bertha a canon used in could fire shots incredibley long distances. The path of a shell could be modeled by y=-0.0196x(squared) plus 1.137x where x was the horizontal distance

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Question 232437: big bertha a canon used in could fire shots incredibley long distances. The path of a shell could be modeled by y=-0.0196x(squared) plus 1.137x where x was the horizontal distance traveled(in miles) and why was the height(in miles).How far could Big Bertha fire a shell? what was the shells maximum height?
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
Given: Y+=+%28-.00196%2AX%5E2+%2B+1.137%2AX%29
Y is the height in miles. When Y = 0, the shell is on the ground.
Solve for Y = 0 and that will give you the distance of the shot.
+0+=+%28-.00196%2AX%5E2+%2B+1.137%2AX%29
Subtract .00196*X^2 from both sides
+.00196%2AX%5E2+=+1.137%2AX
Divide both sides by X
+.00196%2AX+=+1.137
Divide both sides by .00196
X+=+580+miles
So the shell traveled 580 miles
To find the max height of the shell, you need to see where the slope of the line equals 0.
The derivative of the lines equation will give you the slope of the line.
The equation for the line:Y+=+%28-.00196%2AX%5E2+%2B+1.137%2AX%29
Now take the derivative
Y%5E1+=+%282%2A-.00196%2AX+%2B+1.137%29
Set the equation equal to 0 to find the max height.
0+=+%28-.00392%2AX+%2B+1.137%29
Subtract -.00392 from both sides
.00392%2AX+=+1.137%29
Divide both sides by .00392
X+=+290+miles
Now we know that the maximum height occurs when X = 290 miles.
Plug X 290 into the original equation
Y+=+-.00196%2AX%5E2+%2B+1.137%2AX
Y+=+-.00196%2A290%5E2+%2B+1.137%2A290
Y+=+-164.826+%2B+329.73
Y+=+164.90+miles