🚀 go-pugleaf

#>>>> On 11 Jan 93 23:31:18 GMT, mac@cis.ksu.edu (Myron A. Calhoun) said:



mac> "Calculations indicate that the 150-grain .30-caliber Service
mac> bullet fired straight up at a muzzle velocity of 2700 feet per
mac> second will rise 9000 feet, taking about 18 seconds to do it; and
mac> that it takes 31 seconds to return to earth, the last few
mac> thousand feet of the fall being at a nearly constant speed of a
mac> bit over 300 feet per second."

mac> The last clause of the above paragraph just doesn't "compute", at
mac> least not in the way I use the English language!

mac> Suppose "a bit over 300" means "301"; then a constant-speed fall
mac> of 9000 feet at 301 fps takes 29.9 seconds, (and a "nearly
mac> constant speed" shouldn't be much different; see the next two
mac> paragraphs) which doesn't jibe with "the last few thousand feet
mac> of the fall".

if you divide 9000 by 301 you get 29.9. But the bullete doesnt start
at 301 ft/s from 9000 feet.  at 9000 feet the bulletes velocity is 0
ft/s.  it then accelerates at about 30 ft/ss (9.8 m/ss).  then you
have to use this formula
                 distance = .5 * time * time * acceleration

now, due to the air friction, the speed of the bullet doesnt increase
all the time, that's why the author said that for the last few thousand
feet of the fall the velocity is pretty much constant.

if you use this formula and just assum that the speed of the bullet
keeps increasing, then you get time ~ 25s.  but since the bullet will
have a constant speed for a while, 31s seems reasonable.

I'll let the physics majors take it from here!

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