(1) Take derivative, find zero for dy/dx. Plug the x back in to y(x). This finds the max height.Originally Posted byPinky&The Brain

(2) Find zero for y(x). This finds the positions when cannonball is at the ground.

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- Apr 7th 2005, 03:24 PM #1
## Cannonbal Question

Hello,

I am having a hard time with this one, can anybody help?

Cannonball fired on 45 deg angle from horizontal. Initial position is origin. initial velocity 100sqrt(2) ft/sec. Trajectory is y=x-(x/25)^2 where y>=0.

How far till it hits the ground?

What's the max height?

Thanks

- Apr 8th 2005, 02:38 AM #2

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