🚀 go-pugleaf

Myron A. Calhoun (mac@cis.ksu.edu) wrote:
: 
: in his summary of the relevant section (Chapter 20, "Bullets from
: the sky") of Hatcher's Notebook by Julian S. Hatcher, Major General,
: USA, Retired Stackpole Books, Harrisburg, Pennsylvania;
: ISBN 0-8117-0614-1 $24.95 (hardcover)
: jgreely@morganucodon.cis.ohio-state.edu (J Greely) copied
: 
: 	"Calculations indicate that the 150-grain .30-caliber
: 	 Service bullet fired straight up at a muzzle velocity
: 	 of 2700 feet per second will rise 9000 feet, taking
: 	 about 18 seconds to do it; and that it takes 31
: 	 seconds to return to earth, the last few thousand
: 	 feet of the fall being at a nearly constant speed of
: 	 a bit over 300 feet per second."
: 
: The last clause of the above paragraph just doesn't "compute", at least
: not in the way I use the English language!
: 
: Suppose "a bit over 300" means "301";
: then a constant-speed fall of 9000 feet at 301 fps takes 29.9 seconds,
:       (and a "nearly constant speed" shouldn't be
:        much different; see the next two paragraphs)
: which doesn't jibe with "the last few thousand feet of the fall".

I don't see the problem.  Say the last few thousand feet means the last
3000.  This 3000 feet will be traversed in approximately 10 seconds, the
remaining 21 seconds will be used to traverse the previous 6000 feet while
the bullet is accelerating from an initial velocity of 0 to the final
velocity of approx. 300 fps.

Ignoring drag, it would take the bullet approx. 9 seconds to accelerate
from the effect of gravity from 0 to approx. 300 fps so with drag, I find
it easy to believe the 21 second figure.

-Scot