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Math Help - Cannonbal Question

  1. #1
    Newbie pinky&thebrain's Avatar
    Joined
    Mar 2005
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    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
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  2. #2
    Member
    Joined
    Apr 2005
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    Quote Originally Posted by Pinky&The Brain
    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?
    (1) Take derivative, find zero for dy/dx. Plug the x back in to y(x). This finds the max height.

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