SOLUTION: y=x-4000x^2/35739 Using this formula how far downrange does the cannonball travel? What is the maximum height of the cannonball and how far downrange does that height occur?

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Question 949305: y=x-4000x^2/35739
Using this formula how far downrange does the cannonball travel?
What is the maximum height of the cannonball and how far downrange does that height occur?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Apparently y=x-4000x%5E2%2F35739 is the height of a cannonball that has traveled a downrange distance of x , with x and y measured in some unit of length.
The cannonball will travel downrange until it its the ground when y=0 .
That happens when
0=x-4000x%5E2%2F35739<-->0=x%281-4000x%2F35739%29<-->system%28x=0%2C%22or%22%2C1-4000x%2F35739=0%29<-->system%28x=0%2C%22or%22%2Cx=35739%2F4000%29<-->system%28x=0%2C%22or%22%2Cx=8.93475%29 .
That means that the cannonball exits the cannon at ground level, and hits the ground at a distance of highlight%28x=8.93475%29 downrange (in whatever units x is measured).
Halfway to that point (parabolas are symmetrical) the cannonball reaches its maximum height.
Naturally, that happens when
x=%281%2F2%29%2A%2835739%2F4000%29<--->1-4000x%2F35739=1-%284000%2F35739%29%281%2F2%29%2A%2835739%2F4000%29=1-1%2F2=1%2F2 .
Then,
.
Of course, those numbers are exact mathematical calculation results with way to many decimal places. Rounding would be a good idea.
I might say that the cannonball travels downrange a distance of 8.9 units,
and reaches a maximu height of 2.2 units.