Re: how far can a bullet travel shot into water?

Dear David:

Thank you for your question.
The problem you are addressing is an old one and it looks deceivingly 
simple. We need to start mentioning Newton's Second Law. In the following 
link there is a nice explanation about this subject:
 ballistic trajectories

 This page reads in part: "In this lesson we will experiment with 
computing and visualizing ballistic trajectories. A ballistic trajectory 
is the path followed by an object which, after it is given some initial 
velocity, travels only under the influence of gravity. For the purposes of 
this lesson we will ignore the effects of air resistance."

Well, ignoring air resistance, or, more generally, the medium resistance, 
is a huge simplification. Nevertheless, we will go along for the moment. 

The Ballistic Trajectory Problem
One of the first problems studied in an introductory physics class is the 
ballistic trajectory problem. Let's assume that we are standing in a flat 
field and that we throw a ball so that it starts out moving at some 
velocity V > 0 and at some angle theta radians with the ground. At any 
given time t, the ball's horizontal distance is given by 

Vt cos (theta)

and its vertical height is given by 

Vt sin(theta) - (1/2)gt^2

where g is the earth's gravitational constant, or 9.8 meters/sec/sec. This 
assumes that the ball's initial horizontal and vertical positions are both 
0. It will prove convenient to define functions for distance and height 
that take initial velocity, initial angle, and time as arguments and 
return the distance and height of the ball. 

distance := (V, theta, t)     ->      V*t*cos(theta)
height := (V, theta, t)     ->     V*t*sin(theta) - .5*9.8*t^2;

So far, so good. It is clear that if we shot a weapon horizontally (i.e., 
theta = 0), after certain time, the bullet will hit the floor. This is the 
so called range of the weapon.

A nice link presenting interactive graphics is the following:
 http://www.mcasco.com/p1anlm.html

A more formal approach could be find in:
 http://www.phys.uidaho.edu/~pbickers/Course
s/310/Notes/part2/node20.html

Now, since in real life we could not ignore the drag due to the air (or 
water) resistance, we need to refine our ideas.
	It is evident that the drag is a force opposing the movement of 
the bullet. How big is this force?. In
 http:
//www.i3solutions.com/vehiclesimulation/dragequations.htm

 we find:

Air (water) Friction:

Fair = 1/2 Cd A đ V2 

Rolling Resistance:

Rolling = (RRConst + Velocity * RRXcoef) * Weight * Cos(slope)/100 

Grade:

FGrade = Weight*Sin(slope * (22/7)/180)

Total Losses:

FTotal = Fair + FRolling + FGrade 

where Cd is coefficient of drag, A is the cross sectional area, V is the 
velocity, đ is the density of the medium.
It is evident that the bigger the density the shortest distance traveled 
by the bullet. 

Finally, in 
 http://www.las
cruces.com/~jbm/ballistics/secdens.html

we find:

Sectional Density and Ballistic Coefficient 
The BC, or ballistic coefficient is defined as: 
BC = w / [i d2] 
where the diameter is specified in inches and the weight in pounds and the 
form factor is found using: 
i = CD / CDG 
The sectional density is defined as: 
SD = w / d2 
making the ballistic coefficient 
BC = SD / i 
So this means that the ballistic coefficient is proportional to the weight 
of the bullet and inversely proportional to the diameter squared. (Keep in 
mind that the ballistic coefficient is also inversely proportional to the 
form factor which depends on the shape of the bullet!) 
Calculation of the sectional density is straight forward. For a 300 
grain, .338 caliber bullet, the sectional density is: 

SD = [ 300 gr / (7000 gr/lb) ] / [ 0.338 in ]2 = 0.375 lb/in2 

NOTE: With the common definition of the sectional density, the units have 
to be converted when used with drag functions, velocity, etc, to convert 
the in2 to ft2 resulting in a factor of 144. 

Variables 
d 	bullet diameter 	w 	bullet weight 
SD 	sectional density 	BC 	ballistic coefficient 
i 	form factor 	G 	"G" function 
CD 	drag coefficient 	CDG 	drag coefficient of the standard 
bullet

The numerical answer, of course, depends of the particular values of all 
this parameters.

There is a nice book devoted to such problems. It could be found in:
 http://www.border-
barrels.com/book.htm

I hope this helps


Regards


Jaime Valencia