If on it's way upward the bullet is a killing item for sure, on it's way back down earth it's not.
Upward: When the bullet had expended its kinetic E it will stop in the air roughly when E_initial =mgh (h is the height reached _ m the mass of the bullet and g the newtonian accel (=10m/s²)and then fall back
Downward : since initial speed in the fall is null and the mass of the bullet is fairly low the potential energy won't provide a tremendous acceleration by itself and the kinetic E won't regain its initial value as the air will prevent the bullet from reaching tremendous speed by the action of the drag.
if we look at the balance of force on it's way back we hve :
W - D = ma with W the weight (= the mass x newtonian accel) and D the drag (=0.5*ro_air*0.5*Pi*r²*v²)
hence a = g - D/m
As D is a function of the speed, the ratio D/m will increase as the square of the speed gained - -> rapidly toward it's peak value : g - and a (the accel) will be down to zero
If I drop myself out of a plane at 5K my terminal velocity will be around 200kph and then will remain cte until something put an end to this state of balanced E.
This is the same for the bullet. It will fall down at v= 2/r * racine square of (1/10Pi) hence 150 to 180kph (100 to 120mph)
Once it reach someone's skull the potential to hurt would be proportional to the momentum m*v see in regard to the head mass (5kg) hence by the conservation of the quantity m*v
V_head = (m_bullet/m_head)*v_bullet -> the speed of the bullet divided by 500 to 1000 time
it's easy to understand that the E generated would be easily coped by an helmet or the skull itself.
{To make a comparison i's like trowing a little stone weighting 3 time more (average speed 60kph) }
So the impact on the skull would generate less than 0.180kph travel speed what won't penetrate any bone hardened body part (
) .
But it can still hurt !!
~S! and best wishes for the newly freed Libya !
08-23-2011, 10:06 PM
Linear Mode
