The Other Single Bullet Theory

[Charles Collins]

I am not ignoring that at all.  That just gives it the constant rate of drop of v₀sin(2) resulting in a drop of 8.7 feet over 250 feet. The additional drop of 5 feet requires t=sqrt(2h/g)=(2x5/32)^(1/2)=0.56 seconds. That is what limits the speed of the fragment.

Sorry, but I disagree. The dropped bullet accelerates from zero velocity without any other significant force other than gravity. The ricochet has a "running start" due to its momentum. If its velocity is 1500-fps, then I believe that 2/90 = .0222222 X 1500 = 33.33 fps vertically is its "running start."

[Andrew Mason]

Sorry, but I disagree. The dropped bullet accelerates from zero velocity without any other significant force other than gravity. The ricochet has a "running start" due to its momentum. If its velocity is 1500-fps, then I believe that 2/90 = .0222222 X 1500 = 33.33 fps vertically is its "running start."

A projectile follows the following path defined by its horizontal (x) and vertical coordinates (y):

x=x₀ + v₀cos(θ)t
y= y₀ + v₀sin(θ)t - (1/2)gt²

See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5

If you apply that to your situation,
x=x₀=0;
(1)  v₀cos(2)t=250 ft.

y=y₀=0;
v₀sin(2)t=250 sin(2)=8.7 ft
(2)  (1/2)gt²=16t²=5 ft

This last equation (2) means t= √(5/16)=0.56 seconds
Substituting into (1) gives v₀=250/(.56x.9994)=447 fps