Interesting question. Just to get an idea one could go through the following exercise:
Assumptions/Simplifications:
a) Bullet is fired exactly straight upwards.
b) Wind does not matter as it will only move the point of impact but not influence the downward velocity vector.
c) Let’s assume a 158 grain, .357 bullet with a coefficient of drag (Cd) of about 0.8, which is about right for elongated cylindrical objects like a wadcutter. (This is not the BC).
d) Let’s assume the bullet is not tumbling as this would seriously complicate our relatively simple computation.
e) Let’s assume constant, standard atmospheric pressure (even though it would change with altitude).
Based on the formula for terminal velocity
V = Square root ((2*weight)/(Cd*air density*cross sectional area))
the above bullet would travel with about 51 ft/s. Could it penetrate a skull? Don’t know.
FWS
Assumptions/Simplifications:
a) Bullet is fired exactly straight upwards.
b) Wind does not matter as it will only move the point of impact but not influence the downward velocity vector.
c) Let’s assume a 158 grain, .357 bullet with a coefficient of drag (Cd) of about 0.8, which is about right for elongated cylindrical objects like a wadcutter. (This is not the BC).
d) Let’s assume the bullet is not tumbling as this would seriously complicate our relatively simple computation.
e) Let’s assume constant, standard atmospheric pressure (even though it would change with altitude).
Based on the formula for terminal velocity
V = Square root ((2*weight)/(Cd*air density*cross sectional area))
the above bullet would travel with about 51 ft/s. Could it penetrate a skull? Don’t know.
FWS