The bullet will reach one to three miles before falling back to earth.
Basic equation for motion:
D(t) = d + v*t + 1/2*a*t^2
where
t = time (sec)
do = initial height = 0 m
v = velocity of bullet out of gun = 300 m/s (~984 ft/s, based on Philip's info)
a = gravity = -10 (-9.81 m/s^2 really) (gravity)
d = distance as a function of time
Wind resistance ignored.
Motion assumed to be perfectly vertical.
Employ some calculus and solve for maximum distance:
D'(t) = d/dt D(t) = v + a*t
D'(t) = 0 = v + a*t (at max)
v = -a*t (at max)
tmax = -v/a = - 300 / (-10) = 30 sec
Max distance:
D(tmax) = 0 + 300*30 - 1/2*10*30^2 = 4500 m = 4.5 km = ~ 2 miles
Get a freshman or sophomore engineering student in here to check my math. I've not done this for homework in over a decade
Basic equation for motion:
D(t) = d + v*t + 1/2*a*t^2
where
t = time (sec)
do = initial height = 0 m
v = velocity of bullet out of gun = 300 m/s (~984 ft/s, based on Philip's info)
a = gravity = -10 (-9.81 m/s^2 really) (gravity)
d = distance as a function of time
Wind resistance ignored.
Motion assumed to be perfectly vertical.
Employ some calculus and solve for maximum distance:
D'(t) = d/dt D(t) = v + a*t
D'(t) = 0 = v + a*t (at max)
v = -a*t (at max)
tmax = -v/a = - 300 / (-10) = 30 sec
Max distance:
D(tmax) = 0 + 300*30 - 1/2*10*30^2 = 4500 m = 4.5 km = ~ 2 miles
Get a freshman or sophomore engineering student in here to check my math. I've not done this for homework in over a decade