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4 Comments Already

gp4rts Said,
January 4th, 2011 @2:10 pm  

The formula for the distance a projectile will travel fired with initial velocity v and angle ? above horizontal is

d = v²/g * sin2?

the maximum distance is obtained at ? = 45º

g = 9.8 m/s²; v in m/s and d in m.

Steve Said,
January 4th, 2011 @2:25 pm  

The unambiguous equation for level ground is:

R = [Vi²*sin(2?)]/g

Uneven ground is a bit trickier. Email me when you come to that part. I have all the eqs handy.

no ideas < --- Said,
January 4th, 2011 @3:06 pm  

You could set the intial coordinates of the projectile to be at (0,0) and then you can find the horizontal range by using the recipe as follows:

In the horizontal direction the accelertion is zero since the cannon is not dependent on any acceleration in the vertical direction.
dx = v0*cos?*t + 0 + 0

In the vertical direction you do have acceleration so
dy = v0*sin?*t – (1/2)gt^2

When the height of the distance as given by above goes to zero (when it falls to the surface) then you get (1/2)gt^2 = v0*sin?*t and then basically you have the time, so substitute that back into dx and it becomes v0^2 *sin(2?)/g

Philip J Said,
January 4th, 2011 @3:22 pm  

The formula already given is okay for short range artillery where air resistance, coreolis effect and curvature of the Earth are insignificant. If you want to hit a small target a 1000 km away, you’ll need more complex formulas.

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