Re: Playing by the numbers

On Sun, 8 Mar 2009 02:09:30 +0000 (UTC), "Thomas"
<bettablue@xxxxxxxxxxx> wrote:


#I would like to add one more calculation though. I know that gravity is
#always pulling a projectile toward the center of the earth. So what I want
#to do is put together a very simple calculation to show how far the bullet
#will travel before striking the ground. This distance should be figured, for
#my chart, assuming a handgun held at a height of approximately 48 inches. If
#a projectile is traveling at 900 feet /sec, it should theoretically hit the
#ground somewhere in the 300-400 foot area. That is, of course if the gun in
#question were fired level with the ground. I don't want to make it too
#complicated, so I don't want to figure for uphill or downhill shooting.

Distance at a constant acceleration is given by the equation
d=v(0)t+1/2at^2.

Since your example has the gun fired with the barrel level, v(0) in
the y (Vertical) direction is 0. That makes things easy. the
equation becomes:
d=1/2at^2.

Solving for t, one finds that t=sqrt(2d/a). In your example, d=4 ft,
and gravity on this planet ~32.2 ft/s^s.
So, t=sqrt(2*4/32.2) = .498 seconds.

If the velocity of your bullet is 900 ft/s, it will travel about
...498*900=449 ft before it hits the ground.


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